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2022 | 2022_A1 | Given that 0 < |a| < 1, determine the two conditions on b such that the line $y = ax+b$ intersects the curve $y = \ln(1+x^2)$ in exactly one point. Give the final answer as a sum of the expressions b is conditioned on (eg. if the conditions are b < f(a) and b > g(a) then the final solution is f(a) + g(a)). | Write $f(x) = \ln(1+x^2)$. Since the graph of $y=f(x)$ is symmetric under reflection in the $y$-axis, it suffices to consider the case $a \geq 0$: $y=ax+b$ and $y=-ax+b$ intersect $y=f(x)$ the same number of times. For $a=0$, by the symmetry of $y=f(x)$ and the fact that $f(x)> 0$ for all $x\neq 0$ implies that the only line $y=b$ that intersects $y=f(x)$ exactly once is the line $y=0$. We next observe that on $[0,\infty)$, $f'(x) = \frac{2x}{1+x^2}$ increases on $[0,1]$ from $f'(0)=0$ to a maximum at $f'(1)=1$, and then decreases on $[1,\infty)$ with $\lim_{x\to\infty} f'(x)=0$. In particular, $f'(x) \leq 1$ for all $x$ (including $x<0$ since then $f'(x)<0$) and $f'(x)$ achieves each value in $(0,1)$ exactly twice on $[0,\infty)$. For $a \geq 1$, we claim that any line $y=ax+b$ intersects $y=f(x)$ exactly once. They must intersect at least once by the intermediate value theorem: for $x\ll 0$, $ax+b<0<f(x)$, while for $x \gg 0$, $ax+b>f(x)$ since $\lim_{x\to\infty} \frac{\ln(1+x^2)}{x} = 0$. On the other hand, they cannot intersect more than once: for $a>1$, this follows from the mean value theorem, since $f'(x)<a$ for all $x$. For $a=1$, suppose that they intersect at two points $(x_0,y_0)$ and $(x_1,y_1)$. Then \[ 1 = \frac{y_1-y_0}{x_1-x_0} = \frac{\int_{x_0}^{x_1} f'(x)\,dx}{x_1-x_0} < 1 \] since $f'(x)$ is continuous and $f'(x) \leq 1$ with equality only at one point. Finally we consider $0<a<1$. The equation $f'(x) = a$ has exactly two solutions, at $x=r_-$ and $x=r_+$ for $r_{\pm}$ as defined above. If we define $g(x) = f(x)-ax$, then $g'(r_\pm)=0$; $g'$ is strictly decreasing on $(-\infty,r_-)$, strictly increasing on $(r_-,r_+)$, and strictly decreasing on $(r_+,\infty)$; and $\lim_{x\to-\infty} g(x) = \infty$ while $\lim_{x\to\infty} g(x) = -\infty$. It follows that $g(x)=b$ has exactly one solution for $b<g(r_-)$ or $b>g(r_+)$, exactly three solutions for $g(r_-)<b<g(r_+)$, and exactly two solutions for $b = g(r_\pm)$. That is, $y=ax+b$ intersects $y=f(x)$ in exactly one point if and only if $b<g(r_-)$ or $b>g(r_+)$. We thus show that $y=ax+b$ intersects $y=f(x)$ in exactly one point for $0 < |a| < 1$ if and only if, and $b<\ln(1 r_-)^2-|a|r_-$ or $b>\ln(1-r_+)^2-|a|r_+$, where \[r_\pm = \frac{1\pm\sqrt{1-a^2}}{a}.\] thus making the final solution \boxed{2 \ln\left(2 - \frac{2}{a}\right) - 2}. | algebraic | putnam (modified boxing) | Analysis Calculus | Determine all ordered pairs of real numbers $(a,b)$ such that the line $y = ax+b$ intersects the curve $y = \ln(1+x^2)$ in exactly one point. | Write $f(x) = \ln(1+x^2)$. We show that $y=ax+b$ intersects $y=f(x)$ in exactly one point if and only if $(a,b)$ lies in one of the following groups: \begin{itemize} \boxed{\item $a=b=0$ \item $|a| \geq 1$, arbitrary $b$ \item $0 < |a| < 1$, and $b<\ln(1 r_-)^2-|a|r_-$ or $b>\ln(1-r_+)^2-|a|r_+$, where \[r_\pm = \frac{1\pm\sqrt{1-a^2}}{a}.\]} \end{itemize} Since the graph of $y=f(x)$ is symmetric under reflection in the $y$-axis, it suffices to consider the case $a \geq 0$: $y=ax+b$ and $y=-ax+b$ intersect $y=f(x)$ the same number of times. For $a=0$, by the symmetry of $y=f(x)$ and the fact that $f(x)> 0$ for all $x\neq 0$ implies that the only line $y=b$ that intersects $y=f(x)$ exactly once is the line $y=0$. We next observe that on $[0,\infty)$, $f'(x) = \frac{2x}{1+x^2}$ increases on $[0,1]$ from $f'(0)=0$ to a maximum at $f'(1)=1$, and then decreases on $[1,\infty)$ with $\lim_{x\to\infty} f'(x)=0$. In particular, $f'(x) \leq 1$ for all $x$ (including $x<0$ since then $f'(x)<0$) and $f'(x)$ achieves each value in $(0,1)$ exactly twice on $[0,\infty)$. For $a \geq 1$, we claim that any line $y=ax+b$ intersects $y=f(x)$ exactly once. They must intersect at least once by the intermediate value theorem: for $x\ll 0$, $ax+b<0<f(x)$, while for $x \gg 0$, $ax+b>f(x)$ since $\lim_{x\to\infty} \frac{\ln(1+x^2)}{x} = 0$. On the other hand, they cannot intersect more than once: for $a>1$, this follows from the mean value theorem, since $f'(x)<a$ for all $x$. For $a=1$, suppose that they intersect at two points $(x_0,y_0)$ and $(x_1,y_1)$. Then \[ 1 = \frac{y_1-y_0}{x_1-x_0} = \frac{\int_{x_0}^{x_1} f'(x)\,dx}{x_1-x_0} < 1 \] since $f'(x)$ is continuous and $f'(x) \leq 1$ with equality only at one point. Finally we consider $0<a<1$. The equation $f'(x) = a$ has exactly two solutions, at $x=r_-$ and $x=r_+$ for $r_{\pm}$ as defined above. If we define $g(x) = f(x)-ax$, then $g'(r_\pm)=0$; $g'$ is strictly decreasing on $(-\infty,r_-)$, strictly increasing on $(r_-,r_+)$, and strictly decreasing on $(r_+,\infty)$; and $\lim_{x\to-\infty} g(x) = \infty$ while $\lim_{x\to\infty} g(x) = -\infty$. It follows that $g(x)=b$ has exactly one solution for $b<g(r_-)$ or $b>g(r_+)$, exactly three solutions for $g(r_-)<b<g(r_+)$, and exactly two solutions for $b = g(r_\pm)$. That is, $y=ax+b$ intersects $y=f(x)$ in exactly one point if and only if $b<g(r_-)$ or $b>g(r_+)$. | 0 |
2022 | 2022_A2 | Let $n$ be an integer with $n \geq 2$. Over all real polynomials $p(x)$ of degree $n$, what is the largest possible number of negative coefficients of $p(x)^2$? | Write $p(x) = a_nx^n+\cdots+a_1x+a_0$ and $p(x)^2 = b_{2n}x^{2n}+\cdots+b_1x+b_0$. Note that $b_0 = a_0^2$ and $b_{2n} = a_n^2$. We claim that not all of the remaining $2n-1$ coefficients $b_1,\ldots,b_{2n-1}$ can be negative, whence the largest possible number of negative coefficients is $\leq 2n-2$. Indeed, suppose $b_i <0$ for $1\leq i\leq 2n-1$. Since $b_1 = 2a_0a_1$, we have $a_0 \neq 0$. Assume $a_0>0$ (or else replace $p(x)$ by $-p(x)$). We claim by induction on $i$ that $a_i < 0$ for $1\leq i\leq n$. For $i=1$, this follows from $2a_0a_1 = b_1<0$. If $a_i<0$ for $1\leq i\leq k-1$, then \[ 2a_0a_k = b_k - \sum_{i=1}^{k-1} a_i a_{k-i} < b_k < 0 \] and thus $a_k<0$, completing the induction step. But now $b_{2n-1} = 2a_{n-1}a_n > 0$, contradiction. It remains to show that there is a polynomial $p(x)$ such that $p(x)^2$ has $2n-2$ negative coefficients. For example, we may take \[ p(x) = n(x^n+1) - 2(x^{n-1} + \cdots + x), \] so that \begin{align*} p(x)^2 &= n^2(x^{2n} + x^n + 1) - 2n(x^n+1)(x^{n-1}+\cdots+x)\\ &\qquad + (x^{n-1} + \cdots + x)^2. \end{align*} For $i\in \{1,\dots,n-1,n+1,\dots,n-1\}$, the coefficient of $x^i$ in $p(x)^2$ is at most $-2n$ (coming from the cross term) plus $-2n+2$ (from expanding $(x^{n-1} + \cdots + x)^2$), and hence negative. The answer is $\boxed{2n-2}$. | algebraic | putnam | Algebra | Let $n$ be an integer with $n \geq 2$. Over all real polynomials $p(x)$ of degree $n$, what is the largest possible number of negative coefficients of $p(x)^2$? | The answer is $\boxed{2n-2}$. Write $p(x) = a_nx^n+\cdots+a_1x+a_0$ and $p(x)^2 = b_{2n}x^{2n}+\cdots+b_1x+b_0$. Note that $b_0 = a_0^2$ and $b_{2n} = a_n^2$. We claim that not all of the remaining $2n-1$ coefficients $b_1,\ldots,b_{2n-1}$ can be negative, whence the largest possible number of negative coefficients is $\leq 2n-2$. Indeed, suppose $b_i <0$ for $1\leq i\leq 2n-1$. Since $b_1 = 2a_0a_1$, we have $a_0 \neq 0$. Assume $a_0>0$ (or else replace $p(x)$ by $-p(x)$). We claim by induction on $i$ that $a_i < 0$ for $1\leq i\leq n$. For $i=1$, this follows from $2a_0a_1 = b_1<0$. If $a_i<0$ for $1\leq i\leq k-1$, then \[ 2a_0a_k = b_k - \sum_{i=1}^{k-1} a_i a_{k-i} < b_k < 0 \] and thus $a_k<0$, completing the induction step. But now $b_{2n-1} = 2a_{n-1}a_n > 0$, contradiction. It remains to show that there is a polynomial $p(x)$ such that $p(x)^2$ has $2n-2$ negative coefficients. For example, we may take \[ p(x) = n(x^n+1) - 2(x^{n-1} + \cdots + x), \] so that \begin{align*} p(x)^2 &= n^2(x^{2n} + x^n + 1) - 2n(x^n+1)(x^{n-1}+\cdots+x)\\ &\qquad + (x^{n-1} + \cdots + x)^2. \end{align*} For $i\in \{1,\dots,n-1,n+1,\dots,n-1\}$, the coefficient of $x^i$ in $p(x)^2$ is at most $-2n$ (coming from the cross term) plus $-2n+2$ (from expanding $(x^{n-1} + \cdots + x)^2$), and hence negative. | 0 |
2022 | 2022_A3 | Let $p$ be a prime number greater than 5. Let $f(p)$ denote the number of infinite sequences $a_1, a_2, a_3, \dots$ such that $a_n \in \{1, 2, \dots, p-1\}$ and $a_n a_{n+2} \equiv 1 + a_{n+1} \pmod{p}$ for all $n \geq 1$. What is the sum of all values that $f(p)$ can be congruent to modulo 5? | We view the sequence $a_1,a_2,\ldots$ as lying in $\mathbb{F}_p^\times \subset \mathbb{F}_p$. Then the sequence is determined by the values of $a_1$ and $a_2$, via the recurrence $a_{n+2}=(1+a_{n+1})/a_n$. Using this recurrence, we compute \begin{gather*} a_3=\frac{1 + a_2}{a_1}, \, a_4 = \frac{1 + a_1 + a_2}{a_1 a_2}, \\ a_5=\frac{1 + a_1}{a_2}, \, a_6 = a_1, \, a_7 = a_2 \end{gather*} and thus the sequence is periodic with period 5. The values for $a_1$ and $a_2$ may thus be any values in $\mathbb{F}_p^\times$ provided that $a_1\neq p-1$, $a_2\neq p-1$, and $a_1+a_2\neq p-1$. The number of choices for $a_1,a_2\in\{1,\ldots,p-2\}$ such that $a_1+a_2\neq p-1$ is thus $(p-2)^2 - (p-2)= (p-2)(p-3)$. Since $p$ is not a multiple of 5, $(p-2)(p-3)$ is a product of two consecutive integers $a,a+1$, where $a\not\equiv 2 \pmod{5}$. Now $0\cdot 1\equiv 0$, $1\cdot 2 \equiv 2$, $3\cdot 4\equiv 2$, and $4\cdot 0 \equiv 0$ (mod 5). Thus the number of possible sequences $a_1,a_2,\ldots$ is 0 or 2 (mod 5), as desired giving us the final sum to be \boxed{2}. | numerical | putnam (modified boxing) | Combinatorics Number Theory | Let $p$ be a prime number greater than 5. Let $f(p)$ denote the number of infinite sequences $a_1, a_2, a_3, \dots$ such that $a_n \in \{1, 2, \dots, p-1\}$ and $a_n a_{n+2} \equiv 1 + a_{n+1} \pmod{p}$ for all $n \geq 1$. Prove that $f(p)$ is congruent to 0 or 2 $\pmod{5}$. | We view the sequence $a_1,a_2,\ldots$ as lying in $\mathbb{F}_p^\times \subset \mathbb{F}_p$. Then the sequence is determined by the values of $a_1$ and $a_2$, via the recurrence $a_{n+2}=(1+a_{n+1})/a_n$. Using this recurrence, we compute \begin{gather*} a_3=\frac{1 + a_2}{a_1}, \, a_4 = \frac{1 + a_1 + a_2}{a_1 a_2}, \\ a_5=\frac{1 + a_1}{a_2}, \, a_6 = a_1, \, a_7 = a_2 \end{gather*} and thus the sequence is periodic with period 5. The values for $a_1$ and $a_2$ may thus be any values in $\mathbb{F}_p^\times$ provided that $a_1\neq p-1$, $a_2\neq p-1$, and $a_1+a_2\neq p-1$. The number of choices for $a_1,a_2\in\{1,\ldots,p-2\}$ such that $a_1+a_2\neq p-1$ is thus $(p-2)^2 - (p-2)= (p-2)(p-3)$. Since $p$ is not a multiple of 5, $(p-2)(p-3)$ is a product of two consecutive integers $a,a+1$, where $a\not\equiv 2 \pmod{5}$. Now $0\cdot 1\equiv 0$, $1\cdot 2 \equiv 2$, $3\cdot 4\equiv 2$, and $4\cdot 0 \equiv 0$ (mod 5). Thus the number of possible sequences $a_1,a_2,\ldots$ is 0 or 2 (mod 5), as desired. | 0 |
2022 | 2022_A4 | Suppose that $X_1, X_2, \dots$ are real numbers between 0 and 1 that are chosen independently and uniformly at random. Let $S = \sum_{i=1}^k X_i/2^i$, where $k$ is the least positive integer such that $X_k < X_{k+1}$, or $k = \infty$ if there is no such integer. Find the expected value of $S$. | Extend $S$ to an infinite sum by including zero summands for $i> k$. We may then compute the expected value as the sum of the expected value of the $i$-th summand over all $i$. This summand occurs if and only if $X_1,\dots,X_{i-1} \in [X_i, 1]$ and $X_1,\dots,X_{i-1}$ occur in nonincreasing order. These two events are independent and occur with respective probabilities $(1-X_i)^{i-1}$ and $\frac{1}{(i-1)!}$; the expectation of this summand is therefore \begin{align*} &\frac{1}{2^i(i-1)!} \int_0^1 t (1-t)^{i-1}\,dt \\ &\qquad = \frac{1}{2^i(i-1)!} \int_0^1 ((1-t)^{i-1} - (1-t)^i)\,dt \\ &\qquad = \frac{1}{2^i(i-1)!} \left( \frac{1}{i} - \frac{1}{i+1} \right) = \frac{1}{2^i (i+1)!}. \end{align*} Summing over $i$, we obtain \[ \sum_{i=1}^\infty \frac{1}{2^i (i+1)!} = 2 \sum_{i=2}^\infty \frac{1}{2^i i!} = 2\left(e^{1/2}-1-\frac{1}{2} \right). \] The expected value is $\boxed{2e^{1/2}-3}$. | numerical | putnam | Analysis Probability | Suppose that $X_1, X_2, \dots$ are real numbers between 0 and 1 that are chosen independently and uniformly at random. Let $S = \sum_{i=1}^k X_i/2^i$, where $k$ is the least positive integer such that $X_k < X_{k+1}$, or $k = \infty$ if there is no such integer. Find the expected value of $S$. | The expected value is $\boxed{2e^{1/2}-3}$. Extend $S$ to an infinite sum by including zero summands for $i> k$. We may then compute the expected value as the sum of the expected value of the $i$-th summand over all $i$. This summand occurs if and only if $X_1,\dots,X_{i-1} \in [X_i, 1]$ and $X_1,\dots,X_{i-1}$ occur in nonincreasing order. These two events are independent and occur with respective probabilities $(1-X_i)^{i-1}$ and $\frac{1}{(i-1)!}$; the expectation of this summand is therefore \begin{align*} &\frac{1}{2^i(i-1)!} \int_0^1 t (1-t)^{i-1}\,dt \\ &\qquad = \frac{1}{2^i(i-1)!} \int_0^1 ((1-t)^{i-1} - (1-t)^i)\,dt \\ &\qquad = \frac{1}{2^i(i-1)!} \left( \frac{1}{i} - \frac{1}{i+1} \right) = \frac{1}{2^i (i+1)!}. \end{align*} Summing over $i$, we obtain \[ \sum_{i=1}^\infty \frac{1}{2^i (i+1)!} = 2 \sum_{i=2}^\infty \frac{1}{2^i i!} = 2\left(e^{1/2}-1-\frac{1}{2} \right). \] | 0 |
2022 | 2022_A5 | Alice and Bob play a game on a board consisting of one row of 2022 consecutive squares. They take turns placing tiles that cover two adjacent squares, with Alice going first. By rule, a tile must not cover a square that is already covered by another tile. The game ends when no tile can be placed according to this rule. Alice's goal is to maximize the number of uncovered squares when the game ends; Bob's goal is to minimize it. What is the greatest number of uncovered squares that Alice can ensure at the end of the game, no matter how Bob plays? | More generally, let $a(n)$ (resp.\ $b(n)$) be the optimal final score for Alice (resp.\ Bob) moving first in a position with $n$ consecutive squares. We show that \begin{align*} a(n) &= \left\lfloor \frac{n}{7} \right\rfloor + a\left(n - 7\left\lfloor \frac{n}{7} \right\rfloor \right), \\ b(n) &= \left\lfloor \frac{n}{7} \right\rfloor + b\left(n - 7\left\lfloor \frac{n}{7} \right\rfloor \right), \end{align*} and that the values for $n \leq 6$ are as follows: \[ \begin{array}{c|cccccccccc} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline a(n) & 0 & 1 & 0 & 1 & 2 & 1 & 2 \\ b(n) & 0 & 1 & 0 & 1 & 0 & 1 & 0 \end{array} \] Since $2022 \equiv 6 \pmod{7}$, this will yield $a(2022) = 2 + \lfloor \frac{2022}{7} \rfloor = 290$. We proceed by induction, starting with the base cases $n \leq 6$. Since the number of odd intervals never decreases, we have $a(n), b(n) \geq n - 2 \lfloor \frac{n}{2} \rfloor$; by looking at the possible final positions, we see that equality holds for $n=0,1,2,3,5$. For $n=4,6$, Alice moving first can split the original interval into two odd intervals, guaranteeing at least two odd intervals in the final position; whereas Bob can move to leave behind one or two intervals of length 2, guaranteeing no odd intervals in the final position. We now proceed to the induction step. Suppose that $n \geq 7$ and the claim is known for all $m < n$. In particular, this means that $a(m) \geq b(m)$; consequently, it does not change the analysis to allow a player to pass their turn after the first move, as both players will still have an optimal strategy which involves never passing. It will suffice to check that \[ a(n) = a(n-7) + 1, \qquad b(n) = b(n-7) + 1. \] Moving first, Alice can leave behind two intervals of length 1 and $n-3$. This shows that \[ a(n) \geq 1 + b(n-3) = a(n-7) + 1. \] On the other hand, if Alice leaves behind intervals of length $i$ and $n-2-i$, Bob can choose to play in either one of these intervals and then follow Alice's lead thereafter (exercising the pass option if Alice makes the last legal move in one of the intervals). This shows that \begin{align*} a(n) &\leq \max\{\min\{a(i) + b(n-2-i), \\ & \qquad b(i)+a(n-2-i)\}: i =0,1,\dots,n-2\} \\ &= a(n-7)+1. \end{align*} Moving first, Bob can leave behind two intervals of lengths 2 and $n-4$. This shows that \[ b(n) \leq a(n-4) = b(n-7) + 1. \] On the other hand, if Bob leaves behind intervals of length $i$ and $n-2-i$, Alice can choose to play in either one of these intervals and then follow Bob's lead thereafter (again passing as needed). This shows that \begin{align*} b(n) &\geq \min\{\max\{a(i) + b(n-2-i), \\ & \qquad b(i)+a(n-2-i)\}: i =0,1,\dots,n-2\} \\ &= b(n-7)+1. \end{align*} This completes the induction. We show that the number in question equals \boxed{290}. | numerical | putnam | Combinatorics | Alice and Bob play a game on a board consisting of one row of 2022 consecutive squares. They take turns placing tiles that cover two adjacent squares, with Alice going first. By rule, a tile must not cover a square that is already covered by another tile. The game ends when no tile can be placed according to this rule. Alice's goal is to maximize the number of uncovered squares when the game ends; Bob's goal is to minimize it. What is the greatest number of uncovered squares that Alice can ensure at the end of the game, no matter how Bob plays? | We show that the number in question equals \boxed{290}. More generally, let $a(n)$ (resp.\ $b(n)$) be the optimal final score for Alice (resp.\ Bob) moving first in a position with $n$ consecutive squares. We show that \begin{align*} a(n) &= \left\lfloor \frac{n}{7} \right\rfloor + a\left(n - 7\left\lfloor \frac{n}{7} \right\rfloor \right), \\ b(n) &= \left\lfloor \frac{n}{7} \right\rfloor + b\left(n - 7\left\lfloor \frac{n}{7} \right\rfloor \right), \end{align*} and that the values for $n \leq 6$ are as follows: \[ \begin{array}{c|cccccccccc} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline a(n) & 0 & 1 & 0 & 1 & 2 & 1 & 2 \\ b(n) & 0 & 1 & 0 & 1 & 0 & 1 & 0 \end{array} \] Since $2022 \equiv 6 \pmod{7}$, this will yield $a(2022) = 2 + \lfloor \frac{2022}{7} \rfloor = 290$. We proceed by induction, starting with the base cases $n \leq 6$. Since the number of odd intervals never decreases, we have $a(n), b(n) \geq n - 2 \lfloor \frac{n}{2} \rfloor$; by looking at the possible final positions, we see that equality holds for $n=0,1,2,3,5$. For $n=4,6$, Alice moving first can split the original interval into two odd intervals, guaranteeing at least two odd intervals in the final position; whereas Bob can move to leave behind one or two intervals of length 2, guaranteeing no odd intervals in the final position. We now proceed to the induction step. Suppose that $n \geq 7$ and the claim is known for all $m < n$. In particular, this means that $a(m) \geq b(m)$; consequently, it does not change the analysis to allow a player to pass their turn after the first move, as both players will still have an optimal strategy which involves never passing. It will suffice to check that \[ a(n) = a(n-7) + 1, \qquad b(n) = b(n-7) + 1. \] Moving first, Alice can leave behind two intervals of length 1 and $n-3$. This shows that \[ a(n) \geq 1 + b(n-3) = a(n-7) + 1. \] On the other hand, if Alice leaves behind intervals of length $i$ and $n-2-i$, Bob can choose to play in either one of these intervals and then follow Alice's lead thereafter (exercising the pass option if Alice makes the last legal move in one of the intervals). This shows that \begin{align*} a(n) &\leq \max\{\min\{a(i) + b(n-2-i), \\ & \qquad b(i)+a(n-2-i)\}: i =0,1,\dots,n-2\} \\ &= a(n-7)+1. \end{align*} Moving first, Bob can leave behind two intervals of lengths 2 and $n-4$. This shows that \[ b(n) \leq a(n-4) = b(n-7) + 1. \] On the other hand, if Bob leaves behind intervals of length $i$ and $n-2-i$, Alice can choose to play in either one of these intervals and then follow Bob's lead thereafter (again passing as needed). This shows that \begin{align*} b(n) &\geq \min\{\max\{a(i) + b(n-2-i), \\ & \qquad b(i)+a(n-2-i)\}: i =0,1,\dots,n-2\} \\ &= b(n-7)+1. \end{align*} This completes the induction. | 0 |
2022 | 2022_A6 | Let $n$ be a positive integer. Determine, in terms of $n$, the largest integer $m$ with the following property: There exist real numbers $x_1,\dots,x_{2n}$ with $-1 < x_1 < x_2 < \cdots < x_{2n} < 1$ such that the sum of the lengths of the $n$ intervals \[ [x_1^{2k-1}, x_2^{2k-1}], [x_3^{2k-1},x_4^{2k-1}], \dots, [x_{2n-1}^{2k-1}, x_{2n}^{2k-1}] \] is equal to 1 for all integers $k$ with $1 \leq k \leq m$. | To show that $m \geq n$, we take \[ x_j = \cos \frac{(2n+1-j)\pi}{2n+1} \qquad (j=1,\dots,2n). \] It is apparent that $-1 < x_1 < \cdots < x_{2n} < 1$. The sum of the lengths of the intervals can be interpreted as \begin{align*} & -\sum_{j=1}^{2n} ((-1)^{2n+1-j} x_j)^{2k-1} \\ &= -\sum_{j=1}^{2n} \left(\cos (2n+1-j)\left(\pi + \frac{\pi}{2n+1} \right)\right)^{2k-1} \\ &= -\sum_{j=1}^{2n} \left(\cos \frac{2\pi(n+1)j}{2n+1}\right)^{2k-1}. \end{align*} For $\zeta = e^{2 \pi i (n+1)/(2n+1)}$, this becomes \begin{align*} &= -\sum_{j=1}^{2n} \left( \frac{\zeta^j + \zeta^{-j}}{2} \right)^{2k-1} \\ &= -\frac{1}{2^{2k-1}}\sum_{j=1}^{2n} \sum_{l=0}^{2k-1} \binom{2k-1}{l} \zeta^{j(2k-1-2l)} \\ &= -\frac{1}{2^{2k-1}} \sum_{l=0}^{2k-1} \binom{2k-1}{l} \sum_{j=1}^{2n} \zeta^{j(2k-1-2l)} \\ &= -\frac{1}{2^{2k-1}} \sum_{l=0}^{2k-1} \binom{2k-1}{l} (-1) = 1, \end{align*} using the fact that $\zeta^{2k-1-2l}$ is a \emph{nontrivial} root of unity of order dividing $2n+1$. To show that $m \leq n$, we use the following lemma. We say that a multiset $\{x_1,\dots,x_m\}$ of complex numbers is \emph{inverse-free} if there are no two indices $1 \leq i \leq j \leq m$ such that $x_i + x_j = 0$; this implies in particular that 0 does not occur. \begin{lemma*} Let $\{x_1,\dots,x_m\},\{y_1,\dots,y_n\}$ be two inverse-free multisets of complex numbers such that \[ \sum_{i=1}^m x_i^{2k-1} = \sum_{i=1}^n y_i^{2k-1} \qquad (k=1,\dots,\max\{m,n\}). \] Then these two multisets are equal. \end{lemma*} \begin{proof} We may assume without loss of generality that $m \leq n$. Form the rational functions \[ f(z) = \sum_{i=1}^m \frac{x_i z}{1 - x_i^2 z^2}, \quad g(z) = \sum_{i=1}^n \frac{y_i z}{1 - y_i^2 z^2}; \] both $f(z)$ and $g(z)$ have total pole order at most $2n$. Meanwhile, by expanding in power series around $z=0$, we see that $f(z)-g(z)$ is divisible by $z^{2n+1}$. Consequently, the two series are equal. However, we can uniquely recover the multiset $\{x_1,\dots,x_m\}$ from $f(z)$: $f$ has poles at $\{1/x_1^2,\dots,1/x_m^2\}$ and the residue of the pole at $z = 1/x_i^2$ uniquely determines both $x_i$ (i.e., its sign) and its multiplicity. Similarly, we may recover $\{y_1,\dots,y_n\}$ from $g(z)$, so the two multisets must coincide. \end{proof} Now suppose by way of contradiction that we have an example showing that $m \geq n+1$. We then have \[ 1^{2k-1} + \sum_{i=1}^n x_{2i-1}^{2k-1} = \sum_{i=1}^n x_{2i}^{2k-1} \qquad (k=1,\dots,n+1). \] By the lemma, this means that the multisets $\{1,x_1,x_3,\dots,x_{2n-1}\}$ and $\{x_2,x_4,\dots,x_{2n}\}$ become equal after removing pairs of inverses until this becomes impossible. However, of the resulting two multisets, the first contains 1 and the second does not, yielding the desired contradiction. The largest such $m$ is $\boxed{n}$. | algebraic | putnam | Algebra Analysis | Let $n$ be a positive integer. Determine, in terms of $n$, the largest integer $m$ with the following property: There exist real numbers $x_1,\dots,x_{2n}$ with $-1 < x_1 < x_2 < \cdots < x_{2n} < 1$ such that the sum of the lengths of the $n$ intervals \[ [x_1^{2k-1}, x_2^{2k-1}], [x_3^{2k-1},x_4^{2k-1}], \dots, [x_{2n-1}^{2k-1}, x_{2n}^{2k-1}] \] is equal to 1 for all integers $k$ with $1 \leq k \leq m$. | The largest such $m$ is $\boxed{n}$. To show that $m \geq n$, we take \[ x_j = \cos \frac{(2n+1-j)\pi}{2n+1} \qquad (j=1,\dots,2n). \] It is apparent that $-1 < x_1 < \cdots < x_{2n} < 1$. The sum of the lengths of the intervals can be interpreted as \begin{align*} & -\sum_{j=1}^{2n} ((-1)^{2n+1-j} x_j)^{2k-1} \\ &= -\sum_{j=1}^{2n} \left(\cos (2n+1-j)\left(\pi + \frac{\pi}{2n+1} \right)\right)^{2k-1} \\ &= -\sum_{j=1}^{2n} \left(\cos \frac{2\pi(n+1)j}{2n+1}\right)^{2k-1}. \end{align*} For $\zeta = e^{2 \pi i (n+1)/(2n+1)}$, this becomes \begin{align*} &= -\sum_{j=1}^{2n} \left( \frac{\zeta^j + \zeta^{-j}}{2} \right)^{2k-1} \\ &= -\frac{1}{2^{2k-1}}\sum_{j=1}^{2n} \sum_{l=0}^{2k-1} \binom{2k-1}{l} \zeta^{j(2k-1-2l)} \\ &= -\frac{1}{2^{2k-1}} \sum_{l=0}^{2k-1} \binom{2k-1}{l} \sum_{j=1}^{2n} \zeta^{j(2k-1-2l)} \\ &= -\frac{1}{2^{2k-1}} \sum_{l=0}^{2k-1} \binom{2k-1}{l} (-1) = 1, \end{align*} using the fact that $\zeta^{2k-1-2l}$ is a \emph{nontrivial} root of unity of order dividing $2n+1$. To show that $m \leq n$, we use the following lemma. We say that a multiset $\{x_1,\dots,x_m\}$ of complex numbers is \emph{inverse-free} if there are no two indices $1 \leq i \leq j \leq m$ such that $x_i + x_j = 0$; this implies in particular that 0 does not occur. \begin{lemma*} Let $\{x_1,\dots,x_m\},\{y_1,\dots,y_n\}$ be two inverse-free multisets of complex numbers such that \[ \sum_{i=1}^m x_i^{2k-1} = \sum_{i=1}^n y_i^{2k-1} \qquad (k=1,\dots,\max\{m,n\}). \] Then these two multisets are equal. \end{lemma*} \begin{proof} We may assume without loss of generality that $m \leq n$. Form the rational functions \[ f(z) = \sum_{i=1}^m \frac{x_i z}{1 - x_i^2 z^2}, \quad g(z) = \sum_{i=1}^n \frac{y_i z}{1 - y_i^2 z^2}; \] both $f(z)$ and $g(z)$ have total pole order at most $2n$. Meanwhile, by expanding in power series around $z=0$, we see that $f(z)-g(z)$ is divisible by $z^{2n+1}$. Consequently, the two series are equal. However, we can uniquely recover the multiset $\{x_1,\dots,x_m\}$ from $f(z)$: $f$ has poles at $\{1/x_1^2,\dots,1/x_m^2\}$ and the residue of the pole at $z = 1/x_i^2$ uniquely determines both $x_i$ (i.e., its sign) and its multiplicity. Similarly, we may recover $\{y_1,\dots,y_n\}$ from $g(z)$, so the two multisets must coincide. \end{proof} Now suppose by way of contradiction that we have an example showing that $m \geq n+1$. We then have \[ 1^{2k-1} + \sum_{i=1}^n x_{2i-1}^{2k-1} = \sum_{i=1}^n x_{2i}^{2k-1} \qquad (k=1,\dots,n+1). \] By the lemma, this means that the multisets $\{1,x_1,x_3,\dots,x_{2n-1}\}$ and $\{x_2,x_4,\dots,x_{2n}\}$ become equal after removing pairs of inverses until this becomes impossible. However, of the resulting two multisets, the first contains 1 and the second does not, yielding the desired contradiction. | 0 |
2022 | 2022_B1 | Suppose that $P(x) = a_1 x + a_2 x^2 + \cdots + a_n x^n$ is a polynomial with integer coefficients, with $a_1$ odd. Suppose that $e^{P(x)} = b_0 + b_1 x + b_2 x^2 + \cdots$ for all $x$. Find the value of $b_k$ in terms of $k$ and $a_1, a_2, \cdots, a_n$. | We prove that $b_k k!$ is an odd integer for all $k \geq 0$. Since $e^{P(x)} = \sum_{n=0}^\infty \frac{(P(x))^n}{n!}$, the number $k!\,b_k$ is the coefficient of $x^k$ in \[ (P(x))^k + \sum_{n=0}^{k-1} \frac{k!}{n!}(P(x))^n. \] In particular, $b_0=1$ and $b_1=a_1$ are both odd. Now suppose $k \geq 2$; we want to show that $b_k$ is odd. The coefficient of $x^k$ in $(P(x))^k$ is $a_1^k$. It suffices to show that the coefficient of $x^k$ in $\frac{k!}{n!}(P(x))^n$ is an even integer for any $n<k$. For $k$ even or $n \leq k-2$, this follows immediately from the fact that $\frac{k!}{n!}$ is an even integer. For $k$ odd and $n=k-1$, we have \begin{align*} \frac{k!}{(k-1)!}(P(x))^{k-1} &= k(a_1x+a_2x^2+\cdots)^{k-1} \\ &= k(a_1^{k-1}x^{k-1}+(k-1)a_1^{k-2}a_2x^k+\cdots) \end{align*} and the coefficient of $x^k$ is $\boxed{k(k-1)a_1^{k-2}a_2}$, which is again an even integer. | algebraic | putnam (modified boxing) | Algebra Number Theory | Suppose that $P(x) = a_1 x + a_2 x^2 + \cdots + a_n x^n$ is a polynomial with integer coefficients, with $a_1$ odd. Suppose that $e^{P(x)} = b_0 + b_1 x + b_2 x^2 + \cdots$ for all $x$. Prove that $b_k$ is nonzero for all $k \geq 0$. | We prove that $b_k k!$ is an odd integer for all $k \geq 0$. Since $e^{P(x)} = \sum_{n=0}^\infty \frac{(P(x))^n}{n!}$, the number $k!\,b_k$ is the coefficient of $x^k$ in \[ (P(x))^k + \sum_{n=0}^{k-1} \frac{k!}{n!}(P(x))^n. \] In particular, $b_0=1$ and $b_1=a_1$ are both odd. Now suppose $k \geq 2$; we want to show that $b_k$ is odd. The coefficient of $x^k$ in $(P(x))^k$ is $a_1^k$. It suffices to show that the coefficient of $x^k$ in $\frac{k!}{n!}(P(x))^n$ is an even integer for any $n<k$. For $k$ even or $n \leq k-2$, this follows immediately from the fact that $\frac{k!}{n!}$ is an even integer. For $k$ odd and $n=k-1$, we have \begin{align*} \frac{k!}{(k-1)!}(P(x))^{k-1} &= k(a_1x+a_2x^2+\cdots)^{k-1} \\ &= k(a_1^{k-1}x^{k-1}+(k-1)a_1^{k-2}a_2x^k+\cdots) \end{align*} and the coefficient of $x^k$ is $k(k-1)a_1^{k-2}a_2$, which is again an even integer. | 0 |
2022 | 2022_B2 | Let $\times$ represent the cross product in $\mathbb{R}^3$. Find the sum of all positive integers $n$ for which there exist a set $S \subset \mathbb{R}^3$ with exactly $n$ elements such that \[ S = \{v \times w: v, w \in S\}? \] | Clearly the set $S = \{0\}$ works. Suppose that $S \neq \{0\}$ is a finite set satisfying the given condition; in particular, $S$ does not consist of a collection of collinear vectors, since otherwise $\{v \times w: v,w \in S\} = \{0\}$. We claim that $S$ cannot contain any nonzero vector $v$ with $\|v\| \neq 1$. Suppose otherwise, and let $w \in S$ be a vector not collinear with $v$. Then $S$ must contain the nonzero vector $u_1 = v\times w$, as well as the sequence of vectors $u_n$ defined inductively by $u_n = v \times u_{n-1}$. Since each $u_n$ is orthogonal to $v$ by construction, we have $\|u_n\| = \|v\| \|u_{n-1}\|$ and so $\|u_n\| = \|v\|^{n-1} \|u_1\|$. The sequence $\|u_n\|$ consists of all distinct numbers and thus $S$ is infinite, a contradiction. This proves the claim, and so every nonzero vector in $S$ is a unit vector. Next note that any pair of vectors $v,w \in S$ must either be collinear or orthogonal: by the claim, $v,w$ are both unit vectors, and if $v,w$ are not collinear then $v\times w \in S$ must be a unit vector, whence $v\perp w$. Now choose any pair of non-collinear vectors $v_1,v_2 \in S$, and write $v_3 = v_1 \times v_2$. Then $\{v_1,v_2,v_3\}$ is an orthonormal basis of $\mathbb{R}^3$, and it follows that all of these vectors are in $S$: $0$, $v_1$, $v_2$, $v_3$, $-v_1 = v_3 \times v_2$, $-v_2 = v_1 \times v_3$, and $-v_3 = v_2 \times v_1$. On the other hand, $S$ cannot contain any vector besides these seven, since any other vector $w$ in $S$ would have to be simultaneously orthogonal to all of $v_1,v_2,v_3$. Thus any set $S \neq \{0\}$ satisfying the given condition must be of the form $\{0,\pm v_1,\pm v_2,\pm v_3\}$ where $\{v_1,v_2,v_3\}$ is an orthonormal basis of $\mathbb{R}^3$. It is clear that any set of this form does satisfy the given condition. We conclude that the answer is $n=1$ or $n=7$. The possible values of $n$ are 1 and 7, making the sum \boxed{8}. | numerical | putnam (modified boxing) | Linear Algebra | Let $\times$ represent the cross product in $\mathbb{R}^3$. For what positive integers $n$ does there exist a set $S \subset \mathbb{R}^3$ with exactly $n$ elements such that \[ S = \{v \times w: v, w \in S\}? \] | The possible values of $n$ are \boxed{1 and 7}. Clearly the set $S = \{0\}$ works. Suppose that $S \neq \{0\}$ is a finite set satisfying the given condition; in particular, $S$ does not consist of a collection of collinear vectors, since otherwise $\{v \times w: v,w \in S\} = \{0\}$. We claim that $S$ cannot contain any nonzero vector $v$ with $\|v\| \neq 1$. Suppose otherwise, and let $w \in S$ be a vector not collinear with $v$. Then $S$ must contain the nonzero vector $u_1 = v\times w$, as well as the sequence of vectors $u_n$ defined inductively by $u_n = v \times u_{n-1}$. Since each $u_n$ is orthogonal to $v$ by construction, we have $\|u_n\| = \|v\| \|u_{n-1}\|$ and so $\|u_n\| = \|v\|^{n-1} \|u_1\|$. The sequence $\|u_n\|$ consists of all distinct numbers and thus $S$ is infinite, a contradiction. This proves the claim, and so every nonzero vector in $S$ is a unit vector. Next note that any pair of vectors $v,w \in S$ must either be collinear or orthogonal: by the claim, $v,w$ are both unit vectors, and if $v,w$ are not collinear then $v\times w \in S$ must be a unit vector, whence $v\perp w$. Now choose any pair of non-collinear vectors $v_1,v_2 \in S$, and write $v_3 = v_1 \times v_2$. Then $\{v_1,v_2,v_3\}$ is an orthonormal basis of $\mathbb{R}^3$, and it follows that all of these vectors are in $S$: $0$, $v_1$, $v_2$, $v_3$, $-v_1 = v_3 \times v_2$, $-v_2 = v_1 \times v_3$, and $-v_3 = v_2 \times v_1$. On the other hand, $S$ cannot contain any vector besides these seven, since any other vector $w$ in $S$ would have to be simultaneously orthogonal to all of $v_1,v_2,v_3$. Thus any set $S \neq \{0\}$ satisfying the given condition must be of the form $\{0,\pm v_1,\pm v_2,\pm v_3\}$ where $\{v_1,v_2,v_3\}$ is an orthonormal basis of $\mathbb{R}^3$. It is clear that any set of this form does satisfy the given condition. We conclude that the answer is $n=1$ or $n=7$. | 0 |
2022 | 2022_B4 | Find all integers $n$ with $n \geq 4$ for which there exists a sequence of distinct real numbers $x_1,\dots,x_n$ such that each of the sets \begin{gather*} \{x_1,x_2,x_3\}, \{x_2,x_3,x_4\}, \dots, \\ \{x_{n-2},x_{n-1},x_n\}, \{x_{n-1},x_n, x_1\}, \mbox{ and } \{x_n, x_1, x_2\} \end{gather*} forms a 3-term arithmetic progression when arranged in increasing order. Determine the sum of the first $k$ such integers $n$ as a function of $k$. | Note that we interpret ``distinct'' in the problem statement to mean ``pairwise distinct'' (i.e., no two equal). We first show that such a sequence can only occur when $n$ is divisible by 3. If $d_1$ and $d_2$ are the common differences of the arithmetic progressions $\{x_m, x_{m+1}, x_{m+2}\}$ and $\{x_{m+1}, x_{m+2}, x_{m+3}\}$ for some $m$, then $d_2 \in \{d_1, 2d_1, d_1/2\}$. By scaling we may assume that the smallest common difference that occurs is 1; in this case, all of the common differences are integers. By shifting, we may assume that the $x_i$ are themselves all integers. We now observe that any three consecutive terms in the sequence have pairwise distinct residues modulo 3, forcing $n$ to be divisible by 3. We then observe that for any $m \geq 2$, we obtain a sequence of the desired form of length $3m+3 = (2m-1)+1+(m+1)+2$ by concatenating the arithmetic progressions \begin{gather*} (1, 3, \dots, 4m-3, 4m-1), \\ 4m-2, (4m, 4m-4, \dots, 4, 0), 2. \end{gather*} We see that no terms are repeated by noting that the first parenthesized sequence consists of odd numbers; the second sequence consists of multiples of 4; and the remaining numbers $2$ and $4m-2$ are distinct (because $m \geq 2$) but both congruent to 2 mod 4. It remains to show that no such sequence occurs with $n=6$. We may assume without loss of generality that the smallest common difference among the arithmetic progressions is 1 and occurs for $\{x_1, x_2, x_3\}$; by rescaling, shifting, and reversing the sequence as needed, we may assume that $x_1 = 0$ and $(x_2, x_3) \in \{(1,2), (2,1)\}$. We then have $x_4 = 3$ and \[ (x_5, x_6) \in \{(4,5), (-1, -5), (-1, 7), (5, 4), (5, 7)\}. \] In none of these cases does $\{x_5, x_6, 0\}$ form an arithmetic progression. The values of $n$ in question are the multiples of 3 starting with 9, making the required sum of the first k such values \boxed{\frac{3k(k+5)}{2}}. | algebraic | putnam (modified boxing) | Combinatorics Number Theory | Find all integers $n$ with $n \geq 4$ for which there exists a sequence of distinct real numbers $x_1,\dots,x_n$ such that each of the sets \begin{gather*} \{x_1,x_2,x_3\}, \{x_2,x_3,x_4\}, \dots, \\ \{x_{n-2},x_{n-1},x_n\}, \{x_{n-1},x_n, x_1\}, \mbox{ and } \{x_n, x_1, x_2\} \end{gather*} forms a 3-term arithmetic progression when arranged in increasing order. | The values of $n$ in question are the \boxed{multiples of 3 starting with 9}. Note that we interpret ``distinct'' in the problem statement to mean ``pairwise distinct'' (i.e., no two equal). We first show that such a sequence can only occur when $n$ is divisible by 3. If $d_1$ and $d_2$ are the common differences of the arithmetic progressions $\{x_m, x_{m+1}, x_{m+2}\}$ and $\{x_{m+1}, x_{m+2}, x_{m+3}\}$ for some $m$, then $d_2 \in \{d_1, 2d_1, d_1/2\}$. By scaling we may assume that the smallest common difference that occurs is 1; in this case, all of the common differences are integers. By shifting, we may assume that the $x_i$ are themselves all integers. We now observe that any three consecutive terms in the sequence have pairwise distinct residues modulo 3, forcing $n$ to be divisible by 3. We then observe that for any $m \geq 2$, we obtain a sequence of the desired form of length $3m+3 = (2m-1)+1+(m+1)+2$ by concatenating the arithmetic progressions \begin{gather*} (1, 3, \dots, 4m-3, 4m-1), \\ 4m-2, (4m, 4m-4, \dots, 4, 0), 2. \end{gather*} We see that no terms are repeated by noting that the first parenthesized sequence consists of odd numbers; the second sequence consists of multiples of 4; and the remaining numbers $2$ and $4m-2$ are distinct (because $m \geq 2$) but both congruent to 2 mod 4. It remains to show that no such sequence occurs with $n=6$. We may assume without loss of generality that the smallest common difference among the arithmetic progressions is 1 and occurs for $\{x_1, x_2, x_3\}$; by rescaling, shifting, and reversing the sequence as needed, we may assume that $x_1 = 0$ and $(x_2, x_3) \in \{(1,2), (2,1)\}$. We then have $x_4 = 3$ and \[ (x_5, x_6) \in \{(4,5), (-1, -5), (-1, 7), (5, 4), (5, 7)\}. \] In none of these cases does $\{x_5, x_6, 0\}$ form an arithmetic progression. | 0 |
2022 | 2022_B5 | For $0 \leq p \leq 1/2$, let $X_1, X_2, \dots$ be independent random variables such that \[ X_i = \begin{cases} 1 & \mbox{with probability $p$,} \\ -1 & \mbox{with probability $p$,} \\ 0 & \mbox{with probability $1-2p$,} \end{cases} \] for all $i \geq 1$. Given a positive integer $n$ and integers $b, a_1, \dots, a_n$, let $P(b, a_1, \dots, a_n)$ denote the probability that $a_1 X_1 + \cdots + a_n X_n = b$. Find the maximum value of $p$ for which it is the case that \[ P(0, a_1, \dots, a_n) \geq P(b, a_1, \dots, a_n) \] for all positive integers $n$ and all integers $b, a_1, \dots, a_n$? | We first show that $p >1/4$ does not satisfy the desired condition. For $p>1/3$, $P(0,1) = 1-2p < p = P(1,1)$. For $p=1/3$, it is easily calculated (or follows from the next calculation) that $P(0,1,2) = 1/9 < 2/9 = P(1,1,2)$. Now suppose $1/4 < p < 1/3$, and consider $(b,a_1,a_2,a_3,\ldots,a_n) = (1,1,2,4,\ldots,2^{n-1})$. The only solution to \[ X_1+2X_2+\cdots+2^{n-1}X_n = 0 \] with $X_j \in \{0,\pm 1\}$ is $X_1=\cdots=X_n=0$; thus $P(0,1,2,\ldots,2^{2n-1}) = (1-2p)^n$. On the other hand, the solutions to \[ X_1+2X_2+\cdots+2^{n-1}X_n = 1 \] with $X_j \in \{0,\pm 1\}$ are \begin{gather*} (X_1,X_2,\ldots,X_n) = (1,0,\ldots,0),(-1,1,0,\ldots,0), \\ (-1,-1,1,0,\ldots,0), \ldots, (-1,-1,\ldots,-1,1), \end{gather*} and so \begin{align*} &P(1,1,2,\ldots,2^{n-1}) \\ & = p(1-2p)^{n-1}+p^2(1-2p)^{n-2}+\cdots+p^n \\ &= p\frac{(1-2p)^{n}-p^{n}}{1-3p}. \end{align*} It follows that the inequality $P(0,1,2,\ldots,2^{n-1}) \geq P(1,1,2,\ldots,2^{n-1})$ is equivalent to \[ p^{n+1} \geq (4p-1)(1-2p)^n, \] but this is false for sufficiently large $n$ since $4p-1>0$ and $p<1-2p$. Now suppose $p \leq 1/4$; we want to show that for arbitrary $a_1,\ldots,a_n$ and $b \neq 0$, $P(0,a_1,\ldots,a_n) \geq P(b,a_1,\ldots,a_n)$. Define the polynomial \[ f(x) = px+px^{-1}+1-2p, \] and observe that $P(b,a_1,\ldots,a_n)$ is the coefficient of $x^b$ in $f(x^{a_1})f(x^{a_2})\cdots f(x^{a_n})$. We can write \[ f(x^{a_1})f(x^{a_2})\cdots f(x^{a_n}) = g(x)g(x^{-1}) \] for some real polynomial $g$: indeed, if we define $\alpha = \frac{1-2p+\sqrt{1-4p}}{2p} > 0$, then $f(x) = \frac{p}{\alpha}(x+\alpha)(x^{-1}+\alpha)$, and so we can use \[ g(x) = \left(\frac{p}{\alpha}\right)^{n/2} (x^{a_1}+\alpha)\cdots(x^{a_n}+\alpha). \] It now suffices to show that in $g(x)g(x^{-1})$, the coefficient of $x^0$ is at least as large as the coefficient of $x^b$ for any $b \neq 0$. Since $g(x)g(x^{-1})$ is symmetric upon inverting $x$, we may assume that $b > 0$. If we write $g(x) = c_0 x^0 + \cdots + c_m x^m$, then the coefficients of $x^0$ and $x^b$ in $g(x)g(x^{-1})$ are $c_0^2+c_1^2+\cdots+c_m^2$ and $c_0c_b+c_1c_{b+1}+\cdots+c_{m-b}c_m$, respectively. But \begin{align*} &2(c_0c_b+c_1c_{b+1}+\cdots+c_{m-b}c_m)\\ &\leq (c_0^2+c_b^2)+(c_1^2+c_{b+1}^2)+\cdots+(c_{m-b}^2+c_m^2) \\ & \leq 2(c_0^2+\cdots+c_m^2), \end{align*} and the result follows. The answer is $\boxed{1/4}$. | numerical | putnam (modified boxing) | Analysis Probability | For $0 \leq p \leq 1/2$, let $X_1, X_2, \dots$ be independent random variables such that \[ X_i = \begin{cases} 1 & \mbox{with probability $p$,} \\ -1 & \mbox{with probability $p$,} \\ 0 & \mbox{with probability $1-2p$,} \end{cases} \] for all $i \geq 1$. Given a positive integer $n$ and integers $b, a_1, \dots, a_n$, let $P(b, a_1, \dots, a_n)$ denote the probability that $a_1 X_1 + \cdots + a_n X_n = b$. For which values of $p$ is it the case that \[ P(0, a_1, \dots, a_n) \geq P(b, a_1, \dots, a_n) \] for all positive integers $n$ and all integers $b, a_1, \dots, a_n$? | The answer is $\boxed{p \leq 1/4}$. We first show that $p >1/4$ does not satisfy the desired condition. For $p>1/3$, $P(0,1) = 1-2p < p = P(1,1)$. For $p=1/3$, it is easily calculated (or follows from the next calculation) that $P(0,1,2) = 1/9 < 2/9 = P(1,1,2)$. Now suppose $1/4 < p < 1/3$, and consider $(b,a_1,a_2,a_3,\ldots,a_n) = (1,1,2,4,\ldots,2^{n-1})$. The only solution to \[ X_1+2X_2+\cdots+2^{n-1}X_n = 0 \] with $X_j \in \{0,\pm 1\}$ is $X_1=\cdots=X_n=0$; thus $P(0,1,2,\ldots,2^{2n-1}) = (1-2p)^n$. On the other hand, the solutions to \[ X_1+2X_2+\cdots+2^{n-1}X_n = 1 \] with $X_j \in \{0,\pm 1\}$ are \begin{gather*} (X_1,X_2,\ldots,X_n) = (1,0,\ldots,0),(-1,1,0,\ldots,0), \\ (-1,-1,1,0,\ldots,0), \ldots, (-1,-1,\ldots,-1,1), \end{gather*} and so \begin{align*} &P(1,1,2,\ldots,2^{n-1}) \\ & = p(1-2p)^{n-1}+p^2(1-2p)^{n-2}+\cdots+p^n \\ &= p\frac{(1-2p)^{n}-p^{n}}{1-3p}. \end{align*} It follows that the inequality $P(0,1,2,\ldots,2^{n-1}) \geq P(1,1,2,\ldots,2^{n-1})$ is equivalent to \[ p^{n+1} \geq (4p-1)(1-2p)^n, \] but this is false for sufficiently large $n$ since $4p-1>0$ and $p<1-2p$. Now suppose $p \leq 1/4$; we want to show that for arbitrary $a_1,\ldots,a_n$ and $b \neq 0$, $P(0,a_1,\ldots,a_n) \geq P(b,a_1,\ldots,a_n)$. Define the polynomial \[ f(x) = px+px^{-1}+1-2p, \] and observe that $P(b,a_1,\ldots,a_n)$ is the coefficient of $x^b$ in $f(x^{a_1})f(x^{a_2})\cdots f(x^{a_n})$. We can write \[ f(x^{a_1})f(x^{a_2})\cdots f(x^{a_n}) = g(x)g(x^{-1}) \] for some real polynomial $g$: indeed, if we define $\alpha = \frac{1-2p+\sqrt{1-4p}}{2p} > 0$, then $f(x) = \frac{p}{\alpha}(x+\alpha)(x^{-1}+\alpha)$, and so we can use \[ g(x) = \left(\frac{p}{\alpha}\right)^{n/2} (x^{a_1}+\alpha)\cdots(x^{a_n}+\alpha). \] It now suffices to show that in $g(x)g(x^{-1})$, the coefficient of $x^0$ is at least as large as the coefficient of $x^b$ for any $b \neq 0$. Since $g(x)g(x^{-1})$ is symmetric upon inverting $x$, we may assume that $b > 0$. If we write $g(x) = c_0 x^0 + \cdots + c_m x^m$, then the coefficients of $x^0$ and $x^b$ in $g(x)g(x^{-1})$ are $c_0^2+c_1^2+\cdots+c_m^2$ and $c_0c_b+c_1c_{b+1}+\cdots+c_{m-b}c_m$, respectively. But \begin{align*} &2(c_0c_b+c_1c_{b+1}+\cdots+c_{m-b}c_m)\\ &\leq (c_0^2+c_b^2)+(c_1^2+c_{b+1}^2)+\cdots+(c_{m-b}^2+c_m^2) \\ & \leq 2(c_0^2+\cdots+c_m^2), \end{align*} and the result follows. | 0 |
2022 | 2022_B6 | Find all continuous functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that \[ f(xf(y)) + f(yf(x)) = 1 + f(x+y) \] for all $x,y > 0$. Assuming there exists a general expression encompassing all such functions $f$ of the form $f(x) = (ax+b)/(cx+d)$ find b+d. | Note that we interpret $\mathbb{R}^+$ in the problem statement to mean the set of positive real numbers, excluding 0. For convenience, we reproduce here the given equation: \begin{equation} \label{eq:B61} f(xf(y)) + f(yf(x)) = 1 + f(x+y) \end{equation} We first prove that \begin{equation} \label{eq:B62} \lim_{x \to 0^+} f(x) = 1. \end{equation} Set \[ L_- = \liminf_{x \to 0^+} f(x), \quad L_+ = \limsup_{x \to 0^+} f(x). \] For any fixed $y$, we have by \eqref{eq:B61} \begin{align*} L_+ &= \limsup_{x \to 0^+} f(xf(y)) \\ &\leq \limsup_{x \to0^+} (1+f(x+y)) = 1+f(y) < \infty. \end{align*} Consequently, $xf(x) \to 0$ as $x \to 0^+$. By \eqref{eq:B62} with $y=x$, \begin{align*} 2L_+ &= \limsup_{x \to 0^+} 2f(xf(x)) \\ &= \limsup_{x \to 0^+} (1 + f(2x)) = 1 + L_+ \\ 2L_- &= \liminf_{x \to 0^+} 2f(xf(x)) \\ &= \liminf_{x \to 0^+} (1 + f(2x)) = 1 + L_- \end{align*} and so $L_- = L_+ = 1$, confirming \eqref{eq:B62}. We next confirm that \begin{equation} \label{eq:B63} f(x) \geq 1 \mbox{ for all } x>0 \Longrightarrow f(x) = 1 \mbox{ for all } x>0. \end{equation} Suppose that $f(x) \geq 1$ for all $x > 0$. For $0 < c \leq \infty$, put $S_c = \sup\{f(x): 0 < x \leq c\}$; for $c < \infty$, \eqref{eq:B62} implies that $S_c < \infty$. If there exists $y>0$ with $f(y) > 1$, then from \eqref{eq:B61} we have $f(x+y) - f(xf(y)) = f(yf(x)) - 1 \geq 0$; hence \[ S_c = S_{(c-y)f(y)} \qquad \left(c \geq c_0 = \frac{yf(y)}{f(y)-1}\right) \] and (since $(c-y)f(y) - c_0 = f(y)(c-c_0)$) iterating this construction shows that $S_\infty = S_c$ for any $c > c_0$. In any case, we deduce that \begin{equation} \label{eq:B64} f(x) \geq 1 \mbox{ for all } x>0 \Longrightarrow S_\infty < \infty. \end{equation} Still assuming that $f(x) \geq 1$ for all $x>0$, note that from \eqref{eq:B61} with $x=y$, \[ f(xf(x)) = \frac{1}{2}(1 + f(2x)). \] Since $xf(x) \to 0$ as $x \to 0^+$ by \eqref{eq:B62} and $xf(x) \to \infty$ as $x \to \infty$, $xf(x)$ takes all positive real values by the intermediate value theorem. We deduce that $2S_\infty \leq 1 + S_\infty$ and hence $S_\infty = 1$; this proves \eqref{eq:B63}. We may thus assume hereafter that $f(x) < 1$ for some $x > 0$. We next check that \begin{equation} \label{eq:B65} \lim_{x \to \infty} f(x) = 0. \end{equation} Put $I = \inf\{f(x): x > 0\} < 1$, choose $\epsilon \in (0, (1-I)/2)$, and choose $y>0$ such that $f(y) < I+\epsilon$. We then must have $xf(x) \neq y$ for all $x$, or else \[ 1 + I \leq 1 + f(2x) = 2f(y) < 2I + 2\epsilon, \] contradiction. Since $xf(x) \to 0$ as $x \to 0^+$ by \eqref{eq:B62}, we have $\sup\{xf(x): x > 0\} < \infty$ by the intermediate value theorem, yielding \eqref{eq:B65}. By \eqref{eq:B62} plus \eqref{eq:B65}, $f^{-1}(1/2)$ is nonempty and compact. We can now simplify by noting that if $f(x)$ satisfies the original equation, then so does $f(cx)$ for any $c>0$; we may thus assume that the least element of $f^{-1}(1/2)$ is 1, in which case we must show that $f(x) = \frac{1}{1+x}$. We next show that \begin{equation} \label{eq:B68} \lim_{x \to \infty} xf(x) = 1. \end{equation} For all $x > 0$, by \eqref{eq:B61} with $y=x$, \begin{equation} \label{eq:B68a} f(xf(x)) = \frac{1}{2}(1 + f(2x)) > \frac{1}{2} = f(1), \end{equation} so in particular $xf(x) \neq 1$. As in the proof of \eqref{eq:B65}, this implies that $xf(x) < 1$ for all $x > 0$. However, by \eqref{eq:B65} and \eqref{eq:B68a} we have $f(xf(x)) \to \frac{1}{2}$ as $x \to \infty$, yielding \eqref{eq:B68}. By substituting $y \mapsto xy$ in \eqref{eq:B61}, \[ f(xf(xy)) + f(xyf(x)) = 1 + f(x+xy). \] Taking the limit as $x \to \infty$ and applying \eqref{eq:B68} yields \begin{equation} \label{eq:B69} f(1/y) + f(y) = 1. \end{equation} Combining \eqref{eq:B61} with \eqref{eq:B69} yields \[ f(xf(y))=f(x+y)+f \left( \frac{1}{yf(x)} \right). \] Multiply both sides by $xf(y)$, then take the limit as $x \to \infty$ to obtain \begin{align*} 1 &= \lim_{x \to \infty} xf(y) f(x+y) + \lim_{x \to \infty} xf(y) f\left( \frac{1}{yf(x)} \right) \\ &= f(y) + \lim_{x \to \infty} xf(y) yf(x) \\ &= f(y) + yf(y) \end{align*} and solving for $f(y)$ now yields $f(y) = \frac{1}{1+y}$, as desired. The only such functions are the functions $f(x) = \frac{1}{1+cx}$ for some $c \geq 0$ (the case $c=0$ giving the constant function $f(x) = 1$) making the final answer \boxed{2}. | numerical | putnam (modified boxing) | Algebra Analysis | Find all continuous functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that \[ f(xf(y)) + f(yf(x)) = 1 + f(x+y) \] for all $x,y > 0$. | The only such functions are the functions \boxed{$f(x) = \frac{1}{1+cx}$ for some $c \geq 0$ (the case $c=0$ giving the constant function $f(x) = 1$)}. Note that we interpret $\mathbb{R}^+$ in the problem statement to mean the set of positive real numbers, excluding 0. For convenience, we reproduce here the given equation: \begin{equation} \label{eq:B61} f(xf(y)) + f(yf(x)) = 1 + f(x+y) \end{equation} We first prove that \begin{equation} \label{eq:B62} \lim_{x \to 0^+} f(x) = 1. \end{equation} Set \[ L_- = \liminf_{x \to 0^+} f(x), \quad L_+ = \limsup_{x \to 0^+} f(x). \] For any fixed $y$, we have by \eqref{eq:B61} \begin{align*} L_+ &= \limsup_{x \to 0^+} f(xf(y)) \\ &\leq \limsup_{x \to0^+} (1+f(x+y)) = 1+f(y) < \infty. \end{align*} Consequently, $xf(x) \to 0$ as $x \to 0^+$. By \eqref{eq:B62} with $y=x$, \begin{align*} 2L_+ &= \limsup_{x \to 0^+} 2f(xf(x)) \\ &= \limsup_{x \to 0^+} (1 + f(2x)) = 1 + L_+ \\ 2L_- &= \liminf_{x \to 0^+} 2f(xf(x)) \\ &= \liminf_{x \to 0^+} (1 + f(2x)) = 1 + L_- \end{align*} and so $L_- = L_+ = 1$, confirming \eqref{eq:B62}. We next confirm that \begin{equation} \label{eq:B63} f(x) \geq 1 \mbox{ for all } x>0 \Longrightarrow f(x) = 1 \mbox{ for all } x>0. \end{equation} Suppose that $f(x) \geq 1$ for all $x > 0$. For $0 < c \leq \infty$, put $S_c = \sup\{f(x): 0 < x \leq c\}$; for $c < \infty$, \eqref{eq:B62} implies that $S_c < \infty$. If there exists $y>0$ with $f(y) > 1$, then from \eqref{eq:B61} we have $f(x+y) - f(xf(y)) = f(yf(x)) - 1 \geq 0$; hence \[ S_c = S_{(c-y)f(y)} \qquad \left(c \geq c_0 = \frac{yf(y)}{f(y)-1}\right) \] and (since $(c-y)f(y) - c_0 = f(y)(c-c_0)$) iterating this construction shows that $S_\infty = S_c$ for any $c > c_0$. In any case, we deduce that \begin{equation} \label{eq:B64} f(x) \geq 1 \mbox{ for all } x>0 \Longrightarrow S_\infty < \infty. \end{equation} Still assuming that $f(x) \geq 1$ for all $x>0$, note that from \eqref{eq:B61} with $x=y$, \[ f(xf(x)) = \frac{1}{2}(1 + f(2x)). \] Since $xf(x) \to 0$ as $x \to 0^+$ by \eqref{eq:B62} and $xf(x) \to \infty$ as $x \to \infty$, $xf(x)$ takes all positive real values by the intermediate value theorem. We deduce that $2S_\infty \leq 1 + S_\infty$ and hence $S_\infty = 1$; this proves \eqref{eq:B63}. We may thus assume hereafter that $f(x) < 1$ for some $x > 0$. We next check that \begin{equation} \label{eq:B65} \lim_{x \to \infty} f(x) = 0. \end{equation} Put $I = \inf\{f(x): x > 0\} < 1$, choose $\epsilon \in (0, (1-I)/2)$, and choose $y>0$ such that $f(y) < I+\epsilon$. We then must have $xf(x) \neq y$ for all $x$, or else \[ 1 + I \leq 1 + f(2x) = 2f(y) < 2I + 2\epsilon, \] contradiction. Since $xf(x) \to 0$ as $x \to 0^+$ by \eqref{eq:B62}, we have $\sup\{xf(x): x > 0\} < \infty$ by the intermediate value theorem, yielding \eqref{eq:B65}. By \eqref{eq:B62} plus \eqref{eq:B65}, $f^{-1}(1/2)$ is nonempty and compact. We can now simplify by noting that if $f(x)$ satisfies the original equation, then so does $f(cx)$ for any $c>0$; we may thus assume that the least element of $f^{-1}(1/2)$ is 1, in which case we must show that $f(x) = \frac{1}{1+x}$. We next show that \begin{equation} \label{eq:B68} \lim_{x \to \infty} xf(x) = 1. \end{equation} For all $x > 0$, by \eqref{eq:B61} with $y=x$, \begin{equation} \label{eq:B68a} f(xf(x)) = \frac{1}{2}(1 + f(2x)) > \frac{1}{2} = f(1), \end{equation} so in particular $xf(x) \neq 1$. As in the proof of \eqref{eq:B65}, this implies that $xf(x) < 1$ for all $x > 0$. However, by \eqref{eq:B65} and \eqref{eq:B68a} we have $f(xf(x)) \to \frac{1}{2}$ as $x \to \infty$, yielding \eqref{eq:B68}. By substituting $y \mapsto xy$ in \eqref{eq:B61}, \[ f(xf(xy)) + f(xyf(x)) = 1 + f(x+xy). \] Taking the limit as $x \to \infty$ and applying \eqref{eq:B68} yields \begin{equation} \label{eq:B69} f(1/y) + f(y) = 1. \end{equation} Combining \eqref{eq:B61} with \eqref{eq:B69} yields \[ f(xf(y))=f(x+y)+f \left( \frac{1}{yf(x)} \right). \] Multiply both sides by $xf(y)$, then take the limit as $x \to \infty$ to obtain \begin{align*} 1 &= \lim_{x \to \infty} xf(y) f(x+y) + \lim_{x \to \infty} xf(y) f\left( \frac{1}{yf(x)} \right) \\ &= f(y) + \lim_{x \to \infty} xf(y) yf(x) \\ &= f(y) + yf(y) \end{align*} and solving for $f(y)$ now yields $f(y) = \frac{1}{1+y}$, as desired. | 0 |
2023 | 2023_A1 | For a positive integer $n$, let $f_n(x) = \cos(x) \cos(2x) \cos(3x) \cdots \cos(nx)$. Find the smallest $n$ such that $|f_n''(0)| > 2023$. | If we use the product rule to calculate $f_n''(x)$, the result is a sum of terms of two types: terms where two distinct factors $\cos(m_1x)$ and $\cos(m_2x)$ have each been differentiated once, and terms where a single factor $\cos(mx)$ has been differentiated twice. When we evaluate at $x=0$, all terms of the first type vanish since $\sin(0)=0$, while the term of the second type involving $(\cos(mx))''$ becomes $-m^2$. Thus \[ |f_n''(0)| = \left|-\sum_{m=1}^n m^2\right| = \frac{n(n+1)(2n+1)}{6}. \] The function $g(n) = \frac{n(n+1)(2n+1)}{6}$ is increasing for $n\in\mathbb{N}$ and satisfies $g(17)=1785$ and $g(18)=2109$. It follows that the answer is $n=\boxed{18}$. | numerical | putnam | Calculus Trigonometry | For a positive integer $n$, let $f_n(x) = \cos(x) \cos(2x) \cos(3x) \cdots \cos(nx)$. Find the smallest $n$ such that $|f_n''(0)| > 2023$. | If we use the product rule to calculate $f_n''(x)$, the result is a sum of terms of two types: terms where two distinct factors $\cos(m_1x)$ and $\cos(m_2x)$ have each been differentiated once, and terms where a single factor $\cos(mx)$ has been differentiated twice. When we evaluate at $x=0$, all terms of the first type vanish since $\sin(0)=0$, while the term of the second type involving $(\cos(mx))''$ becomes $-m^2$. Thus \[ |f_n''(0)| = \left|-\sum_{m=1}^n m^2\right| = \frac{n(n+1)(2n+1)}{6}. \] The function $g(n) = \frac{n(n+1)(2n+1)}{6}$ is increasing for $n\in\mathbb{N}$ and satisfies $g(17)=1785$ and $g(18)=2109$. It follows that the answer is $n=\boxed{18}$. | 0 |
2023 | 2023_A2 | Let $n$ be an even positive integer. Let $p$ be a monic, real polynomial of degree $2n$; that is to say, $p(x) = x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_1 x + a_0$ for some real coefficients $a_0, \dots, a_{2n-1}$. Suppose that $p(1/k) = k^2$ for all integers $k$ such that $1 \leq |k| \leq n$. Find all other real numbers $x$ for which $p(1/x) = x^2$. | Define the polynomial $q(x) = x^{2n+2}-x^{2n}p(1/x) = x^{2n+2}-(a_0x^{2n}+\cdots+a_{2n-1}x+1)$. The statement that $p(1/x)=x^2$ is equivalent (for $x\neq 0$) to the statement that $x$ is a root of $q(x)$. Thus we know that $\pm 1,\pm 2,\ldots,\pm n$ are roots of $q(x)$, and we can write \[ q(x) = (x^2+ax+b)(x^2-1)(x^2-4)\cdots (x^2-n^2) \] for some monic quadratic polynomial $x^2+ax+b$. Equating the coefficients of $x^{2n+1}$ and $x^0$ on both sides gives $0=a$ and $-1=(-1)^n(n!)^2 b$, respectively. Since $n$ is even, we have $x^2+ax+b = x^2-(n!)^{-2}$. We conclude that there are precisely two other real numbers $x$ such that $p(1/x)=x^2$, and they are $\pm 1/n!$. The only other real numbers with this property are $\boxed{\pm 1/n!}$. (Note that these are indeed \emph{other} values than $\pm 1, \dots, \pm n$ because $n>1$.) | algebraic | putnam | Algebra Number Theory | Let $n$ be an even positive integer. Let $p$ be a monic, real polynomial of degree $2n$; that is to say, $p(x) = x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_1 x + a_0$ for some real coefficients $a_0, \dots, a_{2n-1}$. Suppose that $p(1/k) = k^2$ for all integers $k$ such that $1 \leq |k| \leq n$. Find all other real numbers $x$ for which $p(1/x) = x^2$. | The only other real numbers with this property are $\boxed{\pm 1/n!}$. (Note that these are indeed \emph{other} values than $\pm 1, \dots, \pm n$ because $n>1$.) Define the polynomial $q(x) = x^{2n+2}-x^{2n}p(1/x) = x^{2n+2}-(a_0x^{2n}+\cdots+a_{2n-1}x+1)$. The statement that $p(1/x)=x^2$ is equivalent (for $x\neq 0$) to the statement that $x$ is a root of $q(x)$. Thus we know that $\pm 1,\pm 2,\ldots,\pm n$ are roots of $q(x)$, and we can write \[ q(x) = (x^2+ax+b)(x^2-1)(x^2-4)\cdots (x^2-n^2) \] for some monic quadratic polynomial $x^2+ax+b$. Equating the coefficients of $x^{2n+1}$ and $x^0$ on both sides gives $0=a$ and $-1=(-1)^n(n!)^2 b$, respectively. Since $n$ is even, we have $x^2+ax+b = x^2-(n!)^{-2}$. We conclude that there are precisely two other real numbers $x$ such that $p(1/x)=x^2$, and they are $\pm 1/n!$. | 0 |
2023 | 2023_A3 | Determine the smallest positive real number $r$ such that there exist differentiable functions $f\colon \mathbb{R} \to \mathbb{R}$ and $g\colon \mathbb{R} \to \mathbb{R}$ satisfying \begin{enumerate} \item[(a)] $f(0) > 0$, \item[(b)] $g(0) = 0$, \item[(c)] $|f'(x)| \leq |g(x)|$ for all $x$, \item[(d)] $|g'(x)| \leq |f(x)|$ for all $x$, and \item[(e)] $f(r) = 0$. \end{enumerate} | Suppose by way of contradiction that there exist some $f,g$ satisfying the stated conditions for some $0 < r<\frac{\pi}{2}$. We first note that we can assume that $f(x) \neq 0$ for $x\in [0,r)$. Indeed, by continuity, $\{x\,|\,x\geq 0 \text{ and } f(x)=0\}$ is a closed subset of $[0,\infty)$ and thus has a minimum element $r'$ with $0<r'\leq r$. After replacing $r$ by $r'$, we now have $f(x)\neq 0$ for $x\in [0,r)$. Next we note that $f(r)=0$ implies $g(r) \neq 0$. Indeed, define the function $k :\thinspace \mathbb{R} \to \mathbb{R}$ by $k(x) = f(x)^2+g(x)^2$. Then $|k'(x)| = 2|f(x)f'(x)+g(x)g'(x))| \leq 4|f(x)g(x)| \leq 2k(x)$, where the last inequality follows from the AM-GM inequality. It follows that $\left|\frac{d}{dx} (\log k(x))\right| \leq 2$ for $x \in [0,r)$; since $k(x)$ is continuous at $x=r$, we conclude that $k(r) \neq 0$. Now define the function $h\colon [0,r) \to (-\pi/2,\pi/2)$ by $h(x) = \tan^{-1}(g(x)/f(x))$. We compute that \[ h'(x) = \frac{f(x)g'(x)-g(x)f'(x)}{f(x)^2+g(x)^2} \] and thus \[ |h'(x)| \leq \frac{|f(x)||g'(x)|+|g(x)||f'(x)|}{f(x)^2+g(x)^2} \leq \frac{|f(x)|^2+|g(x)|^2}{f(x)^2+g(x)^2} = 1. \] Since $h(0) = 0$, we have $|h(x)| \leq x<r$ for all $x\in [0,r)$. Since $r<\pi/2$ and $\tan^{-1}$ is increasing on $(-r,r)$, we conclude that $|g(x)/f(x)|$ is uniformly bounded above by $\tan r$ for all $x\in [0,r)$. But this contradicts the fact that $f(r)=0$ and $g(r) \neq 0$, since $\lim_{x\to r^-} g(x)/f(x) = \infty$. This contradiction shows that $r<\pi/2$ cannot be achieved. The answer is $r=\boxed{\frac{\pi}{2}}$, which manifestly is achieved by setting $f(x)=\cos x$ and $g(x)=\sin x$. | numerical | putnam | Analysis Calculus | Determine the smallest positive real number $r$ such that there exist differentiable functions $f\colon \mathbb{R} \to \mathbb{R}$ and $g\colon \mathbb{R} \to \mathbb{R}$ satisfying \begin{enumerate} \item[(a)] $f(0) > 0$, \item[(b)] $g(0) = 0$, \item[(c)] $|f'(x)| \leq |g(x)|$ for all $x$, \item[(d)] $|g'(x)| \leq |f(x)|$ for all $x$, and \item[(e)] $f(r) = 0$. \end{enumerate} | The answer is $r=\boxed{\frac{\pi}{2}}$, which manifestly is achieved by setting $f(x)=\cos x$ and $g(x)=\sin x$. \noindent Suppose by way of contradiction that there exist some $f,g$ satisfying the stated conditions for some $0 < r<\frac{\pi}{2}$. We first note that we can assume that $f(x) \neq 0$ for $x\in [0,r)$. Indeed, by continuity, $\{x\,|\,x\geq 0 \text{ and } f(x)=0\}$ is a closed subset of $[0,\infty)$ and thus has a minimum element $r'$ with $0<r'\leq r$. After replacing $r$ by $r'$, we now have $f(x)\neq 0$ for $x\in [0,r)$. Next we note that $f(r)=0$ implies $g(r) \neq 0$. Indeed, define the function $k :\thinspace \mathbb{R} \to \mathbb{R}$ by $k(x) = f(x)^2+g(x)^2$. Then $|k'(x)| = 2|f(x)f'(x)+g(x)g'(x))| \leq 4|f(x)g(x)| \leq 2k(x)$, where the last inequality follows from the AM-GM inequality. It follows that $\left|\frac{d}{dx} (\log k(x))\right| \leq 2$ for $x \in [0,r)$; since $k(x)$ is continuous at $x=r$, we conclude that $k(r) \neq 0$. Now define the function $h\colon [0,r) \to (-\pi/2,\pi/2)$ by $h(x) = \tan^{-1}(g(x)/f(x))$. We compute that \[ h'(x) = \frac{f(x)g'(x)-g(x)f'(x)}{f(x)^2+g(x)^2} \] and thus \[ |h'(x)| \leq \frac{|f(x)||g'(x)|+|g(x)||f'(x)|}{f(x)^2+g(x)^2} \leq \frac{|f(x)|^2+|g(x)|^2}{f(x)^2+g(x)^2} = 1. \] Since $h(0) = 0$, we have $|h(x)| \leq x<r$ for all $x\in [0,r)$. Since $r<\pi/2$ and $\tan^{-1}$ is increasing on $(-r,r)$, we conclude that $|g(x)/f(x)|$ is uniformly bounded above by $\tan r$ for all $x\in [0,r)$. But this contradicts the fact that $f(r)=0$ and $g(r) \neq 0$, since $\lim_{x\to r^-} g(x)/f(x) = \infty$. This contradiction shows that $r<\pi/2$ cannot be achieved. | 0 |
2023 | 2023_A5 | For a nonnegative integer $k$, let $f(k)$ be the number of ones in the base 3 representation of $k$. Find the sum of all complex numbers $z$ such that \[ \sum_{k=0}^{3^{1010}-1} (-2)^{f(k)} (z+k)^{2023} = 0. \] | We begin by noting that for $n \geq 1$, we have the following equality of polynomials in a parameter $x$: \[ \sum_{k=0}^{3^n-1} (-2)^{f(k)} x^k = \prod_{j=0}^{n-1} (x^{2\cdot 3^j}-2x^{3^j}+1). \] This is readily shown by induction on $n$, using the fact that for $0\leq k\leq 3^{n-1}-1$, $f(3^{n-1}+k)=f(k)+1$ and $f(2\cdot 3^{n-1}+k)=f(k)$. Now define a ``shift'' operator $S$ on polynomials in $z$ by $S(p(z))=p(z+1)$; then we can define $S^m$ for all $m\in\mathbb{Z}$ by $S^m(p(z))$, and in particular $S^0=I$ is the identity map. Write \[ p_n(z) := \sum_{k=0}^{3^n-1}(-2)^{f(k)}(z+k)^{2n+3} \] for $n \geq 1$; it follows that \begin{align*} p_n(z) &= \prod_{j=0}^{n-1}(S^{2\cdot 3^j}-2S^{3^j}+I) z^{2n+3} \\ &= S^{(3^n-1)/2} \prod_{j=0}^{n-1}(S^{3^j}-2I+S^{-3^j}) z^{2n+3}. \end{align*} Next observe that for any $\ell$, the operator $S^\ell-2I+S^{-\ell}$ acts on polynomials in $z$ in a way that decreases degree by $2$. More precisely, for $m\geq 0$, we have \begin{align*} (S^\ell-2I+S^{-\ell})z^m &= (z+\ell)^m-2z^m+(z-\ell)^m \\ &= 2{m\choose 2}\ell^2z^{m-2}+2{m\choose 4}\ell^4z^{m-4}+O(z^{m-6}). \end{align*} We use this general calculation to establish the following: for any $1\leq i\leq n$, there is a nonzero constant $C_i$ (depending on $n$ and $i$ but not $z$) such that \begin{gather} \nonumber \prod_{j=1}^{i} (S^{3^{n-j}}-2I+S^{-3^{n-j}}) z^{2n+3} \\ \nonumber = C_i\left(z^{2n+3-2i}+\textstyle{\frac{(2n+3-2i)(n+1-i)}{6}}(\sum_{j=1}^i 9^{n-j})z^{2n+1-2i}\right) \\ +O(z^{2n-1-2i}). \label{eq:product} \end{gather} Proving \eqref{eq:product} is a straightforward induction on $i$: the induction step applies $S^{3^{n-i-1}}-2I+S^{-3^{n-i-1}}$ to the right hand side of \eqref{eq:product}, using the general formula for $(S^\ell-2I+S^{-\ell})z^m$. Now setting $i=n$ in \eqref{eq:product}, we find that for some $C_n$, \[ \prod_{j=0}^{n-1}(S^{3^j}-2I+S^{-3^j}) z^{2n+3} = C_n\left(z^3+\frac{9^n-1}{16}z\right). \] The roots of this polynomial are $0$ and $\pm \frac{\sqrt{9^n-1}}{4} i$, and it follows that the roots of $p_n(z)$ are these three numbers minus $\frac{3^n-1}{2}$. In particular, when $n=1010$, we find that the roots of $p_{1010}(z)$ are as indicated above. The complex numbers $z$ with this property are \[ -\frac{3^{1010}-1}{2} and -\frac{3^{1010}-1}{2}\pm\frac{\sqrt{9^{1010}-1}}{4}\,i. \] Thus the required sum of all $z$ is \boxed{-\frac{3^{1011}-3}{2}}. | numerical | putnam (modified boxing) | Algebra Complex Numbers | For a nonnegative integer $k$, let $f(k)$ be the number of ones in the base 3 representation of $k$. Find all complex numbers $z$ such that \[ \sum_{k=0}^{3^{1010}-1} (-2)^{f(k)} (z+k)^{2023} = 0. \] | The complex numbers $z$ with this property are \[ \boxed{-\frac{3^{1010}-1}{2} and -\frac{3^{1010}-1}{2}\pm\frac{\sqrt{9^{1010}-1}}{4}\,i}. \] We begin by noting that for $n \geq 1$, we have the following equality of polynomials in a parameter $x$: \[ \sum_{k=0}^{3^n-1} (-2)^{f(k)} x^k = \prod_{j=0}^{n-1} (x^{2\cdot 3^j}-2x^{3^j}+1). \] This is readily shown by induction on $n$, using the fact that for $0\leq k\leq 3^{n-1}-1$, $f(3^{n-1}+k)=f(k)+1$ and $f(2\cdot 3^{n-1}+k)=f(k)$. Now define a ``shift'' operator $S$ on polynomials in $z$ by $S(p(z))=p(z+1)$; then we can define $S^m$ for all $m\in\mathbb{Z}$ by $S^m(p(z))$, and in particular $S^0=I$ is the identity map. Write \[ p_n(z) := \sum_{k=0}^{3^n-1}(-2)^{f(k)}(z+k)^{2n+3} \] for $n \geq 1$; it follows that \begin{align*} p_n(z) &= \prod_{j=0}^{n-1}(S^{2\cdot 3^j}-2S^{3^j}+I) z^{2n+3} \\ &= S^{(3^n-1)/2} \prod_{j=0}^{n-1}(S^{3^j}-2I+S^{-3^j}) z^{2n+3}. \end{align*} Next observe that for any $\ell$, the operator $S^\ell-2I+S^{-\ell}$ acts on polynomials in $z$ in a way that decreases degree by $2$. More precisely, for $m\geq 0$, we have \begin{align*} (S^\ell-2I+S^{-\ell})z^m &= (z+\ell)^m-2z^m+(z-\ell)^m \\ &= 2{m\choose 2}\ell^2z^{m-2}+2{m\choose 4}\ell^4z^{m-4}+O(z^{m-6}). \end{align*} We use this general calculation to establish the following: for any $1\leq i\leq n$, there is a nonzero constant $C_i$ (depending on $n$ and $i$ but not $z$) such that \begin{gather} \nonumber \prod_{j=1}^{i} (S^{3^{n-j}}-2I+S^{-3^{n-j}}) z^{2n+3} \\ \nonumber = C_i\left(z^{2n+3-2i}+\textstyle{\frac{(2n+3-2i)(n+1-i)}{6}}(\sum_{j=1}^i 9^{n-j})z^{2n+1-2i}\right) \\ +O(z^{2n-1-2i}). \label{eq:product} \end{gather} Proving \eqref{eq:product} is a straightforward induction on $i$: the induction step applies $S^{3^{n-i-1}}-2I+S^{-3^{n-i-1}}$ to the right hand side of \eqref{eq:product}, using the general formula for $(S^\ell-2I+S^{-\ell})z^m$. Now setting $i=n$ in \eqref{eq:product}, we find that for some $C_n$, \[ \prod_{j=0}^{n-1}(S^{3^j}-2I+S^{-3^j}) z^{2n+3} = C_n\left(z^3+\frac{9^n-1}{16}z\right). \] The roots of this polynomial are $0$ and $\pm \frac{\sqrt{9^n-1}}{4} i$, and it follows that the roots of $p_n(z)$ are these three numbers minus $\frac{3^n-1}{2}$. In particular, when $n=1010$, we find that the roots of $p_{1010}(z)$ are as indicated above. | 0 |
2023 | 2023_A6 | Alice and Bob play a game in which they take turns choosing integers from $1$ to $n$. Before any integers are chosen, Bob selects a goal of ``odd'' or ``even''. On the first turn, Alice chooses one of the $n$ integers. On the second turn, Bob chooses one of the remaining integers. They continue alternately choosing one of the integers that has not yet been chosen, until the $n$th turn, which is forced and ends the game. Bob wins if the parity of $\{k\colon \mbox{the number $k$ was chosen on the $k$th turn}\}$ matches his goal. Find the sum of all valid values of $n$ for which Bob will have a winning strategy? | Note that we can interpret the game play as building a permutation of $\{1,\dots,n\}$, and the number of times an integer $k$ is chosen on the $k$-th turn is exactly the number of fixed points of this permutation. For $n$ even, Bob selects the goal ``even''. Divide $\{1,\dots,n\}$ into the pairs $\{1,2\},\{3,4\},\dots$; each time Alice chooses an integer, Bob follows suit with the other integer in the same pair. For each pair $\{2k-1,2k\}$, we see that $2k-1$ is a fixed point if and only if $2k$ is, so the number of fixed points is even. For $n$ odd, Bob selects the goal ``odd''. On the first turn, if Alice chooses 1 or 2, then Bob chooses the other one to transpose into the strategy for $n-2$ (with no moves made). We may thus assume hereafter that Alice's first move is some $k > 2$, which Bob counters with 2; at this point there is exactly one fixed point. Thereafter, as long as Alice chooses $j$ on the $j$-th turn (for $j \geq 3$ odd), either $j+1 < k$, in which case Bob can choose $j+1$ to keep the number of fixed points odd; or $j+1=k$, in which case $k$ is even and Bob can choose 1 to transpose into the strategy for $n-k$ (with no moves made). Otherwise, at some odd turn $j$, Alice does not choose $j$. At this point, the number of fixed points is odd, and on each subsequent turn Bob can ensure that neither his own move nor Alice's next move does not create a fixed point: on any turn $j$ for Bob, if $j+1$ is available Bob chooses it; otherwise, Bob has at least two choices available, so he can choose a value other than $j$. For all n, Bob has a winning strategy thus making the required sum $\boxed{n(n+1)/2}$. | algebraic | putnam (modified boxing) | Combinatorics | Alice and Bob play a game in which they take turns choosing integers from $1$ to $n$. Before any integers are chosen, Bob selects a goal of ``odd'' or ``even''. On the first turn, Alice chooses one of the $n$ integers. On the second turn, Bob chooses one of the remaining integers. They continue alternately choosing one of the integers that has not yet been chosen, until the $n$th turn, which is forced and ends the game. Bob wins if the parity of $\{k\colon \mbox{the number $k$ was chosen on the $k$th turn}\}$ matches his goal. For which values of $n$ does Bob have a winning strategy? | (Communicated by Kai Wang) For $\boxed{all n}$, Bob has a winning strategy. Note that we can interpret the game play as building a permutation of $\{1,\dots,n\}$, and the number of times an integer $k$ is chosen on the $k$-th turn is exactly the number of fixed points of this permutation. For $n$ even, Bob selects the goal ``even''. Divide $\{1,\dots,n\}$ into the pairs $\{1,2\},\{3,4\},\dots$; each time Alice chooses an integer, Bob follows suit with the other integer in the same pair. For each pair $\{2k-1,2k\}$, we see that $2k-1$ is a fixed point if and only if $2k$ is, so the number of fixed points is even. For $n$ odd, Bob selects the goal ``odd''. On the first turn, if Alice chooses 1 or 2, then Bob chooses the other one to transpose into the strategy for $n-2$ (with no moves made). We may thus assume hereafter that Alice's first move is some $k > 2$, which Bob counters with 2; at this point there is exactly one fixed point. Thereafter, as long as Alice chooses $j$ on the $j$-th turn (for $j \geq 3$ odd), either $j+1 < k$, in which case Bob can choose $j+1$ to keep the number of fixed points odd; or $j+1=k$, in which case $k$ is even and Bob can choose 1 to transpose into the strategy for $n-k$ (with no moves made). Otherwise, at some odd turn $j$, Alice does not choose $j$. At this point, the number of fixed points is odd, and on each subsequent turn Bob can ensure that neither his own move nor Alice's next move does not create a fixed point: on any turn $j$ for Bob, if $j+1$ is available Bob chooses it; otherwise, Bob has at least two choices available, so he can choose a value other than $j$. | 0 |
2023 | 2023_B1 | Consider an $m$-by-$n$ grid of unit squares, indexed by $(i,j)$ with $1 \leq i \leq m$ and $1 \leq j \leq n$. There are $(m-1)(n-1)$ coins, which are initially placed in the squares $(i,j)$ with $1 \leq i \leq m-1$ and $1 \leq j \leq n-1$. If a coin occupies the square $(i,j)$ with $i \leq m-1$ and $j \leq n-1$ and the squares $(i+1,j), (i,j+1)$, and $(i+1,j+1)$ are unoccupied, then a legal move is to slide the coin from $(i,j)$ to $(i+1,j+1)$. How many distinct configurations of coins can be reached starting from the initial configuration by a (possibly empty) sequence of legal moves? | Initially the unoccupied squares form a path from $(1,n)$ to $(m,1)$ consisting of $m-1$ horizontal steps and $n-1$ vertical steps, and every move preserves this property. This yields an injective map from the set of reachable configurations to the set of paths of this form. Since the number of such paths is evidently $\binom{m+n-2}{m-1}$ (as one can arrange the horizontal and vertical steps in any order), it will suffice to show that the map we just wrote down is also surjective; that is, that one can reach any path of this form by a sequence of moves. This is easiest to see by working backwards. Ending at a given path, if this path is not the initial path, then it contains at least one sequence of squares of the form $(i,j) \to (i,j-1) \to (i+1,j-1)$. In this case the square $(i+1,j)$ must be occupied, so we can undo a move by replacing this sequence with $(i,j) \to (i+1,j) \to (i+1,j-1)$. The number of such configurations is $\boxed{\binom{m+n-2}{m-1}}$. | algebraic | putnam | Algebra Combinatorics | Consider an $m$-by-$n$ grid of unit squares, indexed by $(i,j)$ with $1 \leq i \leq m$ and $1 \leq j \leq n$. There are $(m-1)(n-1)$ coins, which are initially placed in the squares $(i,j)$ with $1 \leq i \leq m-1$ and $1 \leq j \leq n-1$. If a coin occupies the square $(i,j)$ with $i \leq m-1$ and $j \leq n-1$ and the squares $(i+1,j), (i,j+1)$, and $(i+1,j+1)$ are unoccupied, then a legal move is to slide the coin from $(i,j)$ to $(i+1,j+1)$. How many distinct configurations of coins can be reached starting from the initial configuration by a (possibly empty) sequence of legal moves? | The number of such configurations is $\boxed{\binom{m+n-2}{m-1}}$. Initially the unoccupied squares form a path from $(1,n)$ to $(m,1)$ consisting of $m-1$ horizontal steps and $n-1$ vertical steps, and every move preserves this property. This yields an injective map from the set of reachable configurations to the set of paths of this form. Since the number of such paths is evidently $\binom{m+n-2}{m-1}$ (as one can arrange the horizontal and vertical steps in any order), it will suffice to show that the map we just wrote down is also surjective; that is, that one can reach any path of this form by a sequence of moves. This is easiest to see by working backwards. Ending at a given path, if this path is not the initial path, then it contains at least one sequence of squares of the form $(i,j) \to (i,j-1) \to (i+1,j-1)$. In this case the square $(i+1,j)$ must be occupied, so we can undo a move by replacing this sequence with $(i,j) \to (i+1,j) \to (i+1,j-1)$. | 0 |
2023 | 2023_B2 | For each positive integer $n$, let $k(n)$ be the number of ones in the binary representation of $2023 \cdot n$. What is the minimum value of $k(n)$? | We record the factorization $2023 = 7\cdot 17^2$. We first rule out $k(n)=1$ and $k(n)=2$. If $k(n)=1$, then $2023n = 2^a$ for some $a$, which clearly cannot happen. If $k(n)=2$, then $2023n=2^a+2^b=2^b(1+2^{a-b})$ for some $a>b$. Then $1+2^{a-b} \equiv 0\pmod{7}$; but $-1$ is not a power of $2$ mod $7$ since every power of $2$ is congruent to either $1$, $2$, or $4 \pmod{7}$. We now show that there is an $n$ such that $k(n)=3$. It suffices to find $a>b>0$ such that $2023$ divides $2^a+2^b+1$. First note that $2^2+2^1+1=7$ and $2^3 \equiv 1 \pmod{7}$; thus if $a \equiv 2\pmod{3}$ and $b\equiv 1\pmod{3}$ then $7$ divides $2^a+2^b+1$. Next, $2^8+2^5+1 = 17^2$ and $2^{16\cdot 17} \equiv 1 \pmod{17^2}$ by Euler's Theorem; thus if $a \equiv 8 \pmod{16\cdot 17}$ and $b\equiv 5 \pmod{16\cdot 17}$ then $17^2$ divides $2^a+2^b+1$. We have reduced the problem to finding $a,b$ such that $a\equiv 2\pmod{3}$, $a\equiv 8\pmod{16\cdot 17}$, $b\equiv 1\pmod{3}$, $b\equiv 5\pmod{16\cdot 17}$. But by the Chinese Remainder Theorem, integers $a$ and $b$ solving these equations exist and are unique mod $3\cdot 16\cdot 17$. Thus we can find $a,b$ satisfying these congruences; by adding appropriate multiples of $3\cdot 16\cdot 17$, we can also ensure that $a>b>1$. The minimum is $\boxed{3}$. | numerical | putnam | Algebra Number Theory | For each positive integer $n$, let $k(n)$ be the number of ones in the binary representation of $2023 \cdot n$. What is the minimum value of $k(n)$? | The minimum is $\boxed{3}$. \noindent We record the factorization $2023 = 7\cdot 17^2$. We first rule out $k(n)=1$ and $k(n)=2$. If $k(n)=1$, then $2023n = 2^a$ for some $a$, which clearly cannot happen. If $k(n)=2$, then $2023n=2^a+2^b=2^b(1+2^{a-b})$ for some $a>b$. Then $1+2^{a-b} \equiv 0\pmod{7}$; but $-1$ is not a power of $2$ mod $7$ since every power of $2$ is congruent to either $1$, $2$, or $4 \pmod{7}$. We now show that there is an $n$ such that $k(n)=3$. It suffices to find $a>b>0$ such that $2023$ divides $2^a+2^b+1$. First note that $2^2+2^1+1=7$ and $2^3 \equiv 1 \pmod{7}$; thus if $a \equiv 2\pmod{3}$ and $b\equiv 1\pmod{3}$ then $7$ divides $2^a+2^b+1$. Next, $2^8+2^5+1 = 17^2$ and $2^{16\cdot 17} \equiv 1 \pmod{17^2}$ by Euler's Theorem; thus if $a \equiv 8 \pmod{16\cdot 17}$ and $b\equiv 5 \pmod{16\cdot 17}$ then $17^2$ divides $2^a+2^b+1$. We have reduced the problem to finding $a,b$ such that $a\equiv 2\pmod{3}$, $a\equiv 8\pmod{16\cdot 17}$, $b\equiv 1\pmod{3}$, $b\equiv 5\pmod{16\cdot 17}$. But by the Chinese Remainder Theorem, integers $a$ and $b$ solving these equations exist and are unique mod $3\cdot 16\cdot 17$. Thus we can find $a,b$ satisfying these congruences; by adding appropriate multiples of $3\cdot 16\cdot 17$, we can also ensure that $a>b>1$. | 0 |
2023 | 2023_B3 | A sequence $y_1,y_2,\dots,y_k$ of real numbers is called \emph{zigzag} if $k=1$, or if $y_2-y_1, y_3-y_2, \dots, y_k-y_{k-1}$ are nonzero and alternate in sign. Let $X_1,X_2,\dots,X_n$ be chosen independently from the uniform distribution on $[0,1]$. Let $a(X_1,X_2,\dots,X_n)$ be the largest value of $k$ for which there exists an increasing sequence of integers $i_1,i_2,\dots,i_k$ such that $X_{i_1},X_{i_2},\dots,X_{i_k}$ is zigzag. Find the expected value of $a(X_1,X_2,\dots,X_n)$ for $n \geq 2$. | Divide the sequence $X_1,\dots,X_n$ into alternating increasing and decreasing segments, with $N$ segments in all. Note that removing one term cannot increase $N$: if the removed term is interior to some segment then the number remains unchanged, whereas if it separates two segments then one of those decreases in length by 1 (and possibly disappears). From this it follows that $a(X_1,\dots,X_n) = N+1$: in one direction, the endpoints of the segments form a zigzag of length $N+1$; in the other, for any zigzag $X_{i_1},\dots, X_{i_m}$, we can view it as a sequence obtained from $X_1,\dots,X_n$ by removing terms, so its number of segments (which is manifestly $m-1$) cannot exceed $N$. For $n \geq 3$, $a(X_1,\dots,X_n) - a(X_2,\dots,X_{n})$ is 0 if $X_1, X_2, X_3$ form a monotone sequence and 1 otherwise. Since the six possible orderings of $X_1,X_2,X_3$ are equally likely, \[ \mathbf{E}(a(X_1,\dots,X_n) - a(X_1,\dots,X_{n-1})) = \frac{2}{3}. \] Moreover, we always have $a(X_1, X_2) = 2$ because any sequence of two distinct elements is a zigzag. By linearity of expectation plus induction on $n$, we obtain $\mathbf{E}(a(X_1,\dots,X_n)) = \frac{2n+2}{3}$ as claimed. The expected value is $\boxed{\frac{2n+2}{3}}$. | algebraic | putnam | Combinatorics Probability | A sequence $y_1,y_2,\dots,y_k$ of real numbers is called \emph{zigzag} if $k=1$, or if $y_2-y_1, y_3-y_2, \dots, y_k-y_{k-1}$ are nonzero and alternate in sign. Let $X_1,X_2,\dots,X_n$ be chosen independently from the uniform distribution on $[0,1]$. Let $a(X_1,X_2,\dots,X_n)$ be the largest value of $k$ for which there exists an increasing sequence of integers $i_1,i_2,\dots,i_k$ such that $X_{i_1},X_{i_2},\dots,X_{i_k}$ is zigzag. Find the expected value of $a(X_1,X_2,\dots,X_n)$ for $n \geq 2$. | The expected value is $\boxed{\frac{2n+2}{3}}$. Divide the sequence $X_1,\dots,X_n$ into alternating increasing and decreasing segments, with $N$ segments in all. Note that removing one term cannot increase $N$: if the removed term is interior to some segment then the number remains unchanged, whereas if it separates two segments then one of those decreases in length by 1 (and possibly disappears). From this it follows that $a(X_1,\dots,X_n) = N+1$: in one direction, the endpoints of the segments form a zigzag of length $N+1$; in the other, for any zigzag $X_{i_1},\dots, X_{i_m}$, we can view it as a sequence obtained from $X_1,\dots,X_n$ by removing terms, so its number of segments (which is manifestly $m-1$) cannot exceed $N$. For $n \geq 3$, $a(X_1,\dots,X_n) - a(X_2,\dots,X_{n})$ is 0 if $X_1, X_2, X_3$ form a monotone sequence and 1 otherwise. Since the six possible orderings of $X_1,X_2,X_3$ are equally likely, \[ \mathbf{E}(a(X_1,\dots,X_n) - a(X_1,\dots,X_{n-1})) = \frac{2}{3}. \] Moreover, we always have $a(X_1, X_2) = 2$ because any sequence of two distinct elements is a zigzag. By linearity of expectation plus induction on $n$, we obtain $\mathbf{E}(a(X_1,\dots,X_n)) = \frac{2n+2}{3}$ as claimed. | 0 |
2023 | 2023_B4 | For a nonnegative integer $n$ and a strictly increasing sequence of real numbers $t_0,t_1,\dots,t_n$, let $f(t)$ be the corresponding real-valued function defined for $t \geq t_0$ by the following properties: \begin{enumerate} \item[(a)] $f(t)$ is continuous for $t \geq t_0$, and is twice differentiable for all $t>t_0$ other than $t_1,\dots,t_n$; \item[(b)] $f(t_0) = 1/2$; \item[(c)] $\lim_{t \to t_k^+} f'(t) = 0$ for $0 \leq k \leq n$; \item[(d)] For $0 \leq k \leq n-1$, we have $f''(t) = k+1$ when $t_k < t< t_{k+1}$, and $f''(t) = n+1$ when $t>t_n$. \end{enumerate} Considering all choices of $n$ and $t_0,t_1,\dots,t_n$ such that $t_k \geq t_{k-1}+1$ for $1 \leq k \leq n$, what is the least possible value of $T$ for which $f(t_0+T) = 2023$? | Write $t_{n+1} = t_0+T$ and define $s_k = t_k-t_{k-1}$ for $1\leq k\leq n+1$. On $[t_{k-1},t_k]$, we have $f'(t) = k(t-t_{k-1})$ and so $f(t_k)-f(t_{k-1}) = \frac{k}{2} s_k^2$. Thus if we define \[ g(s_1,\ldots,s_{n+1}) = \sum_{k=1}^{n+1} ks_k^2, \] then we want to minimize $\sum_{k=1}^{n+1} s_k = T$ (for all possible values of $n$) subject to the constraints that $g(s_1,\ldots,s_{n+1}) = 4045$ and $s_k \geq 1$ for $k \leq n$. We first note that a minimum value for $T$ is indeed achieved. To see this, note that the constraints $g(s_1,\ldots,s_{n+1}) = 4045$ and $s_k \geq 1$ place an upper bound on $n$. For fixed $n$, the constraint $g(s_1,\ldots,s_{n+1}) = 4045$ places an upper bound on each $s_k$, whence the set of $(s_1,\ldots,s_{n+1})$ on which we want to minimize $\sum s_k$ is a compact subset of $\mathbb{R}^{n+1}$. Now say that $T_0$ is the minimum value of $\sum_{k=1}^{n+1} s_k$ (over all $n$ and $s_1,\ldots,s_{n+1}$), achieved by $(s_1,\ldots,s_{n+1}) = (s_1^0,\ldots,s_{n+1}^0)$. Observe that there cannot be another $(s_1,\ldots,s_{n'+1})$ with the same sum, $\sum_{k=1}^{n'+1} s_k = T_0$, satisfying $g(s_1,\ldots,s_{n'+1}) > 4045$; otherwise, the function $f$ for $(s_1,\ldots,s_{n'+1})$ would satisfy $f(t_0+T_0) > 4045$ and there would be some $T<T_0$ such that $f(t_0+T) = 4045$ by the intermediate value theorem. We claim that $s_{n+1}^0 \geq 1$ and $s_k^0 = 1$ for $1\leq k\leq n$. If $s_{n+1}^0<1$ then \begin{align*} & g(s_1^0,\ldots,s_{n-1}^0,s_n^0+s_{n+1}^0)-g(s_1^0,\ldots,s_{n-1}^0,s_n^0,s_{n+1}^0) \\ &\quad = s_{n+1}^0(2ns_n^0-s_{n+1}^0) > 0, \end{align*} contradicting our observation from the previous paragraph. Thus $s_{n+1}^0 \geq 1$. If $s_k^0>1$ for some $1\leq k\leq n$ then replacing $(s_k^0,s_{n+1}^0)$ by $(1,s_{n+1}^0+s_k^0-1)$ increases $g$: \begin{align*} &g(s_1^0,\ldots,1,\ldots,s_{n+1}^0+s_k^0-1)-g(s_1^0,\ldots,s_k^0,\ldots,s_{n+1}^0) \\ &\quad= (s_k^0-1)((n+1-k)(s_k^0+1)+2(n+1)(s_{n+1}^0-1)) > 0, \end{align*} again contradicting the observation. This establishes the claim. Given that $s_k^0 = 1$ for $1 \leq k \leq n$, we have $T = s_{n+1}^0 + n$ and \[ g(s_1^0,\dots,s_{n+1}^0) = \frac{n(n+1)}{2} + (n+1)(T-n)^2. \] Setting this equal to 4045 and solving for $T$ yields \[ T = n+\sqrt{\frac{4045}{n+1} - \frac{n}{2}}. \] For $n=9$ this yields $T = 29$; it thus suffices to show that for all $n$, \[ n+\sqrt{\frac{4045}{n+1} - \frac{n}{2}} \geq 29. \] This is evident for $n \geq 30$. For $n \leq 29$, rewrite the claim as \[ \sqrt{\frac{4045}{n+1} - \frac{n}{2}} \geq 29-n; \] we then obtain an equivalent inequality by squaring both sides: \[ \frac{4045}{n+1} - \frac{n}{2} \geq n^2-58n+841. \] Clearing denominators, gathering all terms to one side, and factoring puts this in the form \[ (9-n)(n^2 - \frac{95}{2} n + 356) \geq 0. \] The quadratic factor $Q(n)$ has a minimum at $\frac{95}{4} = 23.75$ and satisfies $Q(8) = 40, Q(10) = -19$; it is thus positive for $n \leq 8$ and negative for $10 \leq n \leq 29$. The minimum value of $T$ is $\boxed{29}$. | numerical | putnam | Analysis Calculus | For a nonnegative integer $n$ and a strictly increasing sequence of real numbers $t_0,t_1,\dots,t_n$, let $f(t)$ be the corresponding real-valued function defined for $t \geq t_0$ by the following properties: \begin{enumerate} \item[(a)] $f(t)$ is continuous for $t \geq t_0$, and is twice differentiable for all $t>t_0$ other than $t_1,\dots,t_n$; \item[(b)] $f(t_0) = 1/2$; \item[(c)] $\lim_{t \to t_k^+} f'(t) = 0$ for $0 \leq k \leq n$; \item[(d)] For $0 \leq k \leq n-1$, we have $f''(t) = k+1$ when $t_k < t< t_{k+1}$, and $f''(t) = n+1$ when $t>t_n$. \end{enumerate} Considering all choices of $n$ and $t_0,t_1,\dots,t_n$ such that $t_k \geq t_{k-1}+1$ for $1 \leq k \leq n$, what is the least possible value of $T$ for which $f(t_0+T) = 2023$? | The minimum value of $T$ is $\boxed{29}$. Write $t_{n+1} = t_0+T$ and define $s_k = t_k-t_{k-1}$ for $1\leq k\leq n+1$. On $[t_{k-1},t_k]$, we have $f'(t) = k(t-t_{k-1})$ and so $f(t_k)-f(t_{k-1}) = \frac{k}{2} s_k^2$. Thus if we define \[ g(s_1,\ldots,s_{n+1}) = \sum_{k=1}^{n+1} ks_k^2, \] then we want to minimize $\sum_{k=1}^{n+1} s_k = T$ (for all possible values of $n$) subject to the constraints that $g(s_1,\ldots,s_{n+1}) = 4045$ and $s_k \geq 1$ for $k \leq n$. We first note that a minimum value for $T$ is indeed achieved. To see this, note that the constraints $g(s_1,\ldots,s_{n+1}) = 4045$ and $s_k \geq 1$ place an upper bound on $n$. For fixed $n$, the constraint $g(s_1,\ldots,s_{n+1}) = 4045$ places an upper bound on each $s_k$, whence the set of $(s_1,\ldots,s_{n+1})$ on which we want to minimize $\sum s_k$ is a compact subset of $\mathbb{R}^{n+1}$. Now say that $T_0$ is the minimum value of $\sum_{k=1}^{n+1} s_k$ (over all $n$ and $s_1,\ldots,s_{n+1}$), achieved by $(s_1,\ldots,s_{n+1}) = (s_1^0,\ldots,s_{n+1}^0)$. Observe that there cannot be another $(s_1,\ldots,s_{n'+1})$ with the same sum, $\sum_{k=1}^{n'+1} s_k = T_0$, satisfying $g(s_1,\ldots,s_{n'+1}) > 4045$; otherwise, the function $f$ for $(s_1,\ldots,s_{n'+1})$ would satisfy $f(t_0+T_0) > 4045$ and there would be some $T<T_0$ such that $f(t_0+T) = 4045$ by the intermediate value theorem. We claim that $s_{n+1}^0 \geq 1$ and $s_k^0 = 1$ for $1\leq k\leq n$. If $s_{n+1}^0<1$ then \begin{align*} & g(s_1^0,\ldots,s_{n-1}^0,s_n^0+s_{n+1}^0)-g(s_1^0,\ldots,s_{n-1}^0,s_n^0,s_{n+1}^0) \\ &\quad = s_{n+1}^0(2ns_n^0-s_{n+1}^0) > 0, \end{align*} contradicting our observation from the previous paragraph. Thus $s_{n+1}^0 \geq 1$. If $s_k^0>1$ for some $1\leq k\leq n$ then replacing $(s_k^0,s_{n+1}^0)$ by $(1,s_{n+1}^0+s_k^0-1)$ increases $g$: \begin{align*} &g(s_1^0,\ldots,1,\ldots,s_{n+1}^0+s_k^0-1)-g(s_1^0,\ldots,s_k^0,\ldots,s_{n+1}^0) \\ &\quad= (s_k^0-1)((n+1-k)(s_k^0+1)+2(n+1)(s_{n+1}^0-1)) > 0, \end{align*} again contradicting the observation. This establishes the claim. Given that $s_k^0 = 1$ for $1 \leq k \leq n$, we have $T = s_{n+1}^0 + n$ and \[ g(s_1^0,\dots,s_{n+1}^0) = \frac{n(n+1)}{2} + (n+1)(T-n)^2. \] Setting this equal to 4045 and solving for $T$ yields \[ T = n+\sqrt{\frac{4045}{n+1} - \frac{n}{2}}. \] For $n=9$ this yields $T = 29$; it thus suffices to show that for all $n$, \[ n+\sqrt{\frac{4045}{n+1} - \frac{n}{2}} \geq 29. \] This is evident for $n \geq 30$. For $n \leq 29$, rewrite the claim as \[ \sqrt{\frac{4045}{n+1} - \frac{n}{2}} \geq 29-n; \] we then obtain an equivalent inequality by squaring both sides: \[ \frac{4045}{n+1} - \frac{n}{2} \geq n^2-58n+841. \] Clearing denominators, gathering all terms to one side, and factoring puts this in the form \[ (9-n)(n^2 - \frac{95}{2} n + 356) \geq 0. \] The quadratic factor $Q(n)$ has a minimum at $\frac{95}{4} = 23.75$ and satisfies $Q(8) = 40, Q(10) = -19$; it is thus positive for $n \leq 8$ and negative for $10 \leq n \leq 29$. | 0 |
2023 | 2023_B5 | Determine the sum of the first $k$ positive integers $n$ (in terms of $k$) which have the following property: For all integers $m$ that are relatively prime to $n$, there exists a permutation $\pi\colon \{1,2,\dots,n\} \to \{1,2,\dots,n\}$ such that $\pi(\pi(k)) \equiv mk \pmod{n}$ for all $k \in \{1,2,\dots,n\}$. | Let $\sigma_{n,m}$ be the permutation of $\ZZ/n\ZZ$ induced by multiplication by $m$; the original problem asks for which $n$ does $\sigma_{n,m}$ always have a square root. For $n=1$, $\sigma_{n,m}$ is the identity permutation and hence has a square root. We next identify when a general permutation admits a square root. \begin{lemma} \label{lem:2023B5-2} A permutation $\sigma$ in $S_n$ can be written as the square of another permutation if and only if for every even positive integer $m$, the number of cycles of length $m$ in $\sigma$ is even. \end{lemma} \begin{proof} We first check the ``only if'' direction. Suppose that $\sigma = \tau^2$. Then every cycle of $\tau$ of length $m$ remains a cycle in $\sigma$ if $m$ is odd, and splits into two cycles of length $m/2$ if $m$ is even. We next check the ``if'' direction. We may partition the cycles of $\sigma$ into individual cycles of odd length and pairs of cycles of the same even length; then we may argue as above to write each partition as the square of another permutation. \end{proof} Suppose now that $n>1$ is odd. Write $n = p^e k$ where $p$ is an odd prime, $k$ is a positive integer, and $\gcd(p,k) = 1$. By the Chinese remainder theorem, we have a ring isomorphism \[ \ZZ/n\ZZ \cong \ZZ/p^e \ZZ \times \ZZ/k \ZZ. \] Recall that the group $(\ZZ/p^e \ZZ)^\times$ is cyclic; choose $m \in \ZZ$ reducing to a generator of $(\ZZ/p^e \ZZ)^\times$ and to the identity in $(\ZZ/k\ZZ)^\times$. Then $\sigma_{n,m}$ consists of $k$ cycles (an odd number) of length $p^{e-1}(p-1)$ (an even number) plus some shorter cycles. By Lemma~\ref{lem:2023B5-2}, $\sigma_{n,m}$ does not have a square root. Suppose next that $n \equiv 2 \pmod{4}$. Write $n = 2k$ with $k$ odd, so that \[ \ZZ/n\ZZ \cong \ZZ/2\ZZ \times \ZZ/k\ZZ. \] Then $\sigma_{n,m}$ acts on $\{0\} \times \ZZ/k\ZZ$ and $\{1\} \times \ZZ/k\ZZ$ with the same cycle structure, so every cycle length occurs an even number of times. By Lemma~\ref{lem:2023B5-2}, $\sigma_{n,m}$ has a square root. Finally, suppose that $n$ is divisible by 4. For $m = -1$, $\sigma_{n,m}$ consists of two fixed points ($0$ and $n/2$) together with $n/2-1$ cycles (an odd number) of length 2 (an even number). By Lemma~\ref{lem:2023B5-2}, $\sigma_{n,m}$ does not have a square root. The desired property holds if and only if n = 1 or $n \equiv 2 \pmod{4}$, hence making the required sum $\boxed{2k^2-4k+3}$. | algebraic | putnam (modified boxing) | Combinatorics Number Theory | Determine which positive integers $n$ have the following property: For all integers $m$ that are relatively prime to $n$, there exists a permutation $\pi\colon \{1,2,\dots,n\} \to \{1,2,\dots,n\}$ such that $\pi(\pi(k)) \equiv mk \pmod{n}$ for all $k \in \{1,2,\dots,n\}$. | The desired property holds if and only if $\boxed{n = 1 or n \equiv 2 \pmod{4}}$. Let $\sigma_{n,m}$ be the permutation of $\ZZ/n\ZZ$ induced by multiplication by $m$; the original problem asks for which $n$ does $\sigma_{n,m}$ always have a square root. For $n=1$, $\sigma_{n,m}$ is the identity permutation and hence has a square root. We next identify when a general permutation admits a square root. \begin{lemma} \label{lem:2023B5-2} A permutation $\sigma$ in $S_n$ can be written as the square of another permutation if and only if for every even positive integer $m$, the number of cycles of length $m$ in $\sigma$ is even. \end{lemma} \begin{proof} We first check the ``only if'' direction. Suppose that $\sigma = \tau^2$. Then every cycle of $\tau$ of length $m$ remains a cycle in $\sigma$ if $m$ is odd, and splits into two cycles of length $m/2$ if $m$ is even. We next check the ``if'' direction. We may partition the cycles of $\sigma$ into individual cycles of odd length and pairs of cycles of the same even length; then we may argue as above to write each partition as the square of another permutation. \end{proof} Suppose now that $n>1$ is odd. Write $n = p^e k$ where $p$ is an odd prime, $k$ is a positive integer, and $\gcd(p,k) = 1$. By the Chinese remainder theorem, we have a ring isomorphism \[ \ZZ/n\ZZ \cong \ZZ/p^e \ZZ \times \ZZ/k \ZZ. \] Recall that the group $(\ZZ/p^e \ZZ)^\times$ is cyclic; choose $m \in \ZZ$ reducing to a generator of $(\ZZ/p^e \ZZ)^\times$ and to the identity in $(\ZZ/k\ZZ)^\times$. Then $\sigma_{n,m}$ consists of $k$ cycles (an odd number) of length $p^{e-1}(p-1)$ (an even number) plus some shorter cycles. By Lemma~\ref{lem:2023B5-2}, $\sigma_{n,m}$ does not have a square root. Suppose next that $n \equiv 2 \pmod{4}$. Write $n = 2k$ with $k$ odd, so that \[ \ZZ/n\ZZ \cong \ZZ/2\ZZ \times \ZZ/k\ZZ. \] Then $\sigma_{n,m}$ acts on $\{0\} \times \ZZ/k\ZZ$ and $\{1\} \times \ZZ/k\ZZ$ with the same cycle structure, so every cycle length occurs an even number of times. By Lemma~\ref{lem:2023B5-2}, $\sigma_{n,m}$ has a square root. Finally, suppose that $n$ is divisible by 4. For $m = -1$, $\sigma_{n,m}$ consists of two fixed points ($0$ and $n/2$) together with $n/2-1$ cycles (an odd number) of length 2 (an even number). By Lemma~\ref{lem:2023B5-2}, $\sigma_{n,m}$ does not have a square root. | 0 |
2023 | 2023_B6 | Let $n$ be a positive integer. For $i$ and $j$ in $\{1,2,\dots,n\}$, let $s(i,j)$ be the number of pairs $(a,b)$ of nonnegative integers satisfying $ai +bj=n$. Let $S$ be the $n$-by-$n$ matrix whose $(i,j)$ entry is $s(i,j)$. For example, when $n=5$, we have $S = \begin{bmatrix} 6 & 3 & 2 & 2 & 2 \\ 3 & 0 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 & 1 \\ 2 & 0 & 0 & 0 & 1 \\ 2 & 1 & 1 & 1 & 2 \end{bmatrix}$. Compute the determinant of $S$. \end{itemize} \end{document} | To begin with, we read off the following features of $S$. \begin{itemize} \item $S$ is symmetric: $S_{ij} = S_{ji}$ for all $i,j$, corresponding to $(a,b) \mapsto (b,a)$). \item $S_{11} = n+1$, corresponding to $(a,b) = (0,n),(1,n-1),\dots,(n,0)$. \item If $n = 2m$ is even, then $S_{mj} = 3$ for $j=1,m$, corresponding to $(a,b) = (2,0),(1,\frac{n}{2j}),(0,\frac{n}{j})$. \item For $\frac{n}{2} < i \leq n$, $S_{ij} = \# (\ZZ \cap \{\frac{n-i}{j}, \frac{n}{j}\})$, corresponding to $(a,b) = (1, \frac{n-i}{j}), (0, \frac{n}{j})$. \end{itemize} Let $T$ be the matrix obtained from $S$ by performing row and column operations as follows: for $d=2,\dots,n-2$, subtract $S_{nd}$ times row $n-1$ from row $d$ and subtract $S_{nd}$ times column $n-1$ from column $d$; then subtract row $n-1$ from row $n$ and column $n-1$ from column $n$. Evidently $T$ is again symmetric and $\det(T) = \det(S)$. Let us examine row $i$ of $T$ for $\frac{n}{2} < i < n-1$: \begin{align*} T_{i1} &= S_{i1} - S_{in} S_{(n-1)1} = 2-1\cdot 2 = 0 \\ T_{ij} &= S_{ij} - S_{in} S_{(n-1)j} - S_{nj}S_{i(n-1)}\\ & = \begin{cases} 1 & \mbox{if $j$ divides $n-i$} \\ 0 & \mbox{otherwise}. \end{cases} \quad (1 < j < n-1) \\ T_{i(n-1)} &= S_{i(n-1)} - S_{in} S_{(n-1)(n-1)} = 0-1\cdot0 = 0 \\ T_{in} &= S_{in} - S_{in} S_{(n-1)n} - S_{i(n-1)} = 1 - 1\cdot1 - 0 = 0. \end{align*} Now recall (e.g., from the expansion of a determinant in minors) if a matrix contains an entry equal to 1 which is the unique nonzero entry in either its row or its column, then we may strike out this entry (meaning striking out the row and column containing it) at the expense of multiplying the determinant by a sign. To simplify notation, we do \emph{not} renumber rows and columns after performing this operation. We next verify that for the matrix $T$, for $i=2,\dots,\lfloor \frac{n}{2} \rfloor$ in turn, it is valid to strike out $(i,n-i)$ and $(n-i, i)$ at the cost of multiplying the determinant by -1. Namely, when we reach the entry $(n-i,i)$, the only other nonzero entries in this row have the form $(n-i,j)$ where $j>1$ divides $n-i$, and those entries are in previously struck columns. We thus compute $\det(S) = \det(T)$ as: \begin{gather*} (-1)^{\lfloor n/2 \rfloor-1} \det \begin{pmatrix} n+1 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \mbox{for $n$ odd,} \\ (-1)^{\lfloor n/2 \rfloor-1} \det \begin{pmatrix} n+1 & -1 & 2 & 0 \\ -1 & -1 & 1 & -1 \\ 2 & 1 & 0 & 1 \\ 0 & -1 & 1 & 0 \end{pmatrix} \mbox{for $n$ even.} \end{gather*} In the odd case, we can strike the last two rows and columns (creating another negation) and then conclude at once. In the even case, the rows and columns are labeled $1, \frac{n}{2}, n-1, n$; by adding row/column $n-1$ to row/column $\frac{n}{2}$, we produce \[ (-1)^{\lfloor n/2 \rfloor} \det \begin{pmatrix} n+1 & 1 & 2 & 0 \\ 1 & 1 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \] and we can again strike the last two rows and columns (creating another negation) and then read off the result. The determinant equals $\boxed{(-1)^{\lceil n/2 \rceil-1} 2 \lceil \frac{n}{2} \rceil}$. | algebraic | putnam | Combinatorics Linear Algebra | Let $n$ be a positive integer. For $i$ and $j$ in $\{1,2,\dots,n\}$, let $s(i,j)$ be the number of pairs $(a,b)$ of nonnegative integers satisfying $ai +bj=n$. Let $S$ be the $n$-by-$n$ matrix whose $(i,j)$ entry is $s(i,j)$. For example, when $n=5$, we have $S = \begin{bmatrix} 6 & 3 & 2 & 2 & 2 \\ 3 & 0 & 1 & 0 & 1 \\ 2 & 1 & 0 & 0 & 1 \\ 2 & 0 & 0 & 0 & 1 \\ 2 & 1 & 1 & 1 & 2 \end{bmatrix}$. Compute the determinant of $S$. \end{itemize} \end{document} | The determinant equals $\boxed{(-1)^{\lceil n/2 \rceil-1} 2 \lceil \frac{n}{2} \rceil}$. To begin with, we read off the following features of $S$. \begin{itemize} \item $S$ is symmetric: $S_{ij} = S_{ji}$ for all $i,j$, corresponding to $(a,b) \mapsto (b,a)$). \item $S_{11} = n+1$, corresponding to $(a,b) = (0,n),(1,n-1),\dots,(n,0)$. \item If $n = 2m$ is even, then $S_{mj} = 3$ for $j=1,m$, corresponding to $(a,b) = (2,0),(1,\frac{n}{2j}),(0,\frac{n}{j})$. \item For $\frac{n}{2} < i \leq n$, $S_{ij} = \# (\ZZ \cap \{\frac{n-i}{j}, \frac{n}{j}\})$, corresponding to $(a,b) = (1, \frac{n-i}{j}), (0, \frac{n}{j})$. \end{itemize} Let $T$ be the matrix obtained from $S$ by performing row and column operations as follows: for $d=2,\dots,n-2$, subtract $S_{nd}$ times row $n-1$ from row $d$ and subtract $S_{nd}$ times column $n-1$ from column $d$; then subtract row $n-1$ from row $n$ and column $n-1$ from column $n$. Evidently $T$ is again symmetric and $\det(T) = \det(S)$. Let us examine row $i$ of $T$ for $\frac{n}{2} < i < n-1$: \begin{align*} T_{i1} &= S_{i1} - S_{in} S_{(n-1)1} = 2-1\cdot 2 = 0 \\ T_{ij} &= S_{ij} - S_{in} S_{(n-1)j} - S_{nj}S_{i(n-1)}\\ & = \begin{cases} 1 & \mbox{if $j$ divides $n-i$} \\ 0 & \mbox{otherwise}. \end{cases} \quad (1 < j < n-1) \\ T_{i(n-1)} &= S_{i(n-1)} - S_{in} S_{(n-1)(n-1)} = 0-1\cdot0 = 0 \\ T_{in} &= S_{in} - S_{in} S_{(n-1)n} - S_{i(n-1)} = 1 - 1\cdot1 - 0 = 0. \end{align*} Now recall (e.g., from the expansion of a determinant in minors) if a matrix contains an entry equal to 1 which is the unique nonzero entry in either its row or its column, then we may strike out this entry (meaning striking out the row and column containing it) at the expense of multiplying the determinant by a sign. To simplify notation, we do \emph{not} renumber rows and columns after performing this operation. We next verify that for the matrix $T$, for $i=2,\dots,\lfloor \frac{n}{2} \rfloor$ in turn, it is valid to strike out $(i,n-i)$ and $(n-i, i)$ at the cost of multiplying the determinant by -1. Namely, when we reach the entry $(n-i,i)$, the only other nonzero entries in this row have the form $(n-i,j)$ where $j>1$ divides $n-i$, and those entries are in previously struck columns. We thus compute $\det(S) = \det(T)$ as: \begin{gather*} (-1)^{\lfloor n/2 \rfloor-1} \det \begin{pmatrix} n+1 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \mbox{for $n$ odd,} \\ (-1)^{\lfloor n/2 \rfloor-1} \det \begin{pmatrix} n+1 & -1 & 2 & 0 \\ -1 & -1 & 1 & -1 \\ 2 & 1 & 0 & 1 \\ 0 & -1 & 1 & 0 \end{pmatrix} \mbox{for $n$ even.} \end{gather*} In the odd case, we can strike the last two rows and columns (creating another negation) and then conclude at once. In the even case, the rows and columns are labeled $1, \frac{n}{2}, n-1, n$; by adding row/column $n-1$ to row/column $\frac{n}{2}$, we produce \[ (-1)^{\lfloor n/2 \rfloor} \det \begin{pmatrix} n+1 & 1 & 2 & 0 \\ 1 & 1 & 1 & 0 \\ 2 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} \] and we can again strike the last two rows and columns (creating another negation) and then read off the result. | 0 |
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