IMO Shortlist
Sequence
combinatorics
inequality system
induction

$ a_{kn \plus{} i} \equal{} 0$ if $ 0 \leq i \leq n \minus{} k$
and
$ a_{kn \plus{} i} \equal{} 1$ if $ n \minus{} k < i \leq n$

Let $ b_i \equal{} a_{i\cdot n \plus{} 1} \plus{} a_{i \cdot n \plus{} 2} \plus{} \cdots \plus{} a_{(i \plus{} 1)n}$
then $ 0 \leq b_0 < b_1 < \cdots b_n \leq n \Rightarrow b_i \equal{} i$

suppose $ a_{s n \plus{} t}$ is greatest element satisfy $ a_{s n \plus{} t}$ such that $ a_{sn \plus{} t} \equal{} 0$ and $ n \minus{} s < t \leq n$

Let $ c_i \equal{} a_{i \cdot n \plus{} 1} \plus{} a_{i \cdot n \plus{} 2} \plus{} \cdots \plus{} a_{i \cdot n \plus{} t \minus{} 1}$
and $ d_i \equal{} a_{i \cdot n \plus{} t} \plus{} a_{i \cdot n \plus{} t \plus{} 1} \plus{} \cdots \plus{} a_{(i \plus{} 1)n}$

$ 0 \equal{} d_0 \plus{} c_1 < d_1 \plus{} c_2 < \cdots d_s \plus{} c_{s \plus{} 1} \equal{} s \minus{} 1 \Rightarrow d_i \plus{} c_{i \plus{} 1} \equal{} i (i \leq s)$

$ b_i \minus{} (d_i \plus{} c_{i \plus{} 1}) \equal{} c_i \plus{} d_i \minus{} d_i \minus{} c_{i \plus{} 1} \equal{} 0 \Rightarrow c_i \equal{} c_{i \plus{} 1} (i \leq s)$

because $ c_0 \leq b_0 \Rightarrow c_0 \equal{} 0 \Rightarrow c_s \equal{} 0$
But $ d_s \leq s \minus{} 1$ because $ a_{sn \plus{} t} \equal{} 0$ then $ b_s \equal{} c_s \plus{} d_s \leq s \minus{} 1$ Contradiction!!

I have the exact same solution as nayel :). It's nice.

We induct on $n\ge1$ to show that $a_{kn+i}=1$ ($0\le k\le n$ and $1\le i\le n$) iff $k\ge i$, where the base case is clear.

Let $s_i=a_i+\cdots+a_{i+n-1}$. Note that $0\le s_i\le n$ for all $0\le i\le n^2+1$. Since
\[s_1 < s_{n+1} < \cdots < s_{n^2+1},\]we have $s_{kn+1}=k$ for $0\le k\le n$. In particular,
\[a_1=\cdots=a_n=0\wedge a_{n^2+1}+\cdots+a_{n^2+n}=1.\]Thus
\begin{align*}
A = s_2+s_{n+2}+\cdots+s_{(n-1)n+2} &= s_{n+1}+s_{2n+1}+\cdots+s_{(n-1)n+1}+1 \\
&= 1+2+\cdots+(n-1)+1 = B.
\end{align*}However, $s_2<s_{n+2}<\cdots<s_{(n-1)n+2}$ gives us $s_{kn+2}\in[k,k+1]$ for $0\le k\le n-1$. If $s_{(n-1)n+2}=n-1$, though, we must have
\[A = s_2+s_{n+2}+\cdots+s_{(n-1)n+2} \le 1+2+\cdots+(n-1) < B,\]a contradiction. So $s_{(n-1)n+2}=n$, whence
\[B-n = 1+2+\cdots+(n-2) \ge s_2+s_{n+2}+\cdots+s_{(n-2)n+2} = A-s_{(n-1)n+2} = A-n\]and so $s_{kn+2}=k=s_{kn+1}$ for $0\le k\le n-2$. Letting $k$ range from $0$ to $n-2$, we find that
\[0=a_1=a_{n+1}=\cdots=a_{(n-1)n+1}.\]But $s_{(n-1)n+1}=n-1$, so
\[a_{(n-1)n+2}=\cdots=a_{(n-1)n+n}=1.\]
It's now easy to see that $a_{kn+i}$, for $0\le k\le n-1$ and $2\le i\le n$ satisfy the inductive hypothesis, so we're done. (Visualizing the numbers in a $(n+1)\times n$ matrix also helps.)

Can someone read my solution and give feedback on how to structure/word it better? Thanks :)
[hide=Solution][b]Lemma 1:[/b] There are $r$ 1's from $rn+1$ to $(r+1)n$
There are $n+1$ intervals, each one is between $0$ and $n$, and each one must be greater than the previous. Hence, they must be $0,1,\dots,n$. The lemma follows.

I claim that the only sequence is the sequence such that $a_{rn+i}$, where $1\le i \le n$, is on if and only if $i>n-r$, where $i<n$. For example, the only sequence for $n=5$ would be
\[00000\text{ }00001\text{ }00011\text{ }00111\text{ }01111\text{ }11111\]

We prove the result with induction on $r$. This is trivially true for our base case of $r=0$ since everything is $0$ by lemma 1. Now, supposing the result is true for $r-1$, we wish to prove the result for $r$. Suppose for the sake of contradiction that $a_{rn+i}$ is the first $1$ from $rn+1$ to $(r+1)n$ and $i \le n-r$. Now, we consider the interval from $(r-1)n+i+1$ to $rn+i$. Let $x_{i,j}$ denote the number of ones from $i \cdot n+1$ to $i \cdot n+j$. By our inductive hypothesis and our assumption, this interval has $(r-1)+1=r$ 1s. In particular, $x_{r,i}=1$.

Now, we claim that $x_{r+j,i} \ge j+1$.We prove this using induction on $j$, our base case is already true since $x_{r,i}=1$. Suppose the result is true for some $j$. By lemma 1 there are $r+j$ 1's from $(r+j)n+1$ to $(r+j+1)n$. By our inductive hypothesis, $x_{r+j,i} \ge j+1$. Hence, there are at most $(r+j)-(j+1)=r-1$ 1's from $(r+j)n+i+1$ to $(r+j+1)n$. In order to satisfy the second condition, there must be at least $(r+j)-(r-1)+1=j+2$ ones from $(r+j+1)n+1$ to $(r+j+1)n+i$. Hence, $x_{r+j+1,i} \ge j+2$ as desired, completing our induction.

Now, take $j=n-r$. We have that $x_{n,i}=n-r+1$. However, by our assumption, $i\le n-r$, so we cannot have a sum of $n-r+1$ in $n-r$ places since everything is 0's and 1's by the first condition. Hence, we get a contradiction, and $i>n-r$. On the other hand, by lemma 1, there are exactly $r$ 1s from $rn+1$ to $(r+1)n$, so they must occupy the last $r$ places: $rn-r+1, \dots (r+1)n$. This also completes our induction on $r$.

Hence, there is exactly one sequence, namely the on such that $a_{rn+i}$, where $1\le i \le n$, is on if and only if $i>n-r$, where $i<n$.[/hide]
EDIT: Sorry about the weird formatting, I don't know of a good way to split formulas in half other than doing it manually

Felt more like an Algebra problem to me. Annoying indexing, wow. There must be an easier solution than mine, though, lol.

We claim that the only possible sequence is \begin{align*} a_1=a_2=\dots=a_n &=0
\\ a_{n+1}=a_{n+2}=\dots=a_{2n-1}=0, \ a_{2n}&=1
\\ a_{2n+1}=a_{2n+2}=\dots=a_{3n-2}=0, \ a_{3n-1}=a_{3n} &=1
\\ &\dots
\\ a_{n^2}=a_{n^2+1}=\dots=a_{n^2+n} &=1
\end{align*} Note the chain of inequalities \[ 0 \leq a_1+\dots+a_n < a_{n+1}+\dots+a_{2n} < \dots < a_{n^2+1}+\dots+a_{n^2+n} \leq n. \] As those sums are all integers and there are exactly $n$ strict 'less than'-signs, we can conclude $\sum_{i=1}^{n} a_{kn+i} = k$ for $0 \leq k \leq n$. Furthermore, note \[ 0 \leq \sum_{i=j}^{n+j-1} a_i < \sum_{i=j}^{n+j-1} a_{n+i} < \dots < \sum_{i=j}^{n+j-1} a_{n^2+i} \leq n \] for $2 \leq j \leq n$. Now those are exactly $n-1$ strict inequality signs. Thus $k \leq \sum_{i=j}^{n+j-1} a_{kn+i} \leq k+1$. Now we'll induct.
[b]Base Case:[/b] Notice $a_1=a_2=\dots=a_n=0$, as $\sum_{i=1}^{n} a_i=0$.
[b]Induction Hypothesis:[/b] Let $(a_{kn+1},a_{kn+2},\dots,a_{(k+1)n}) = (0,0,\dots,0,1,1,\dots,1)$ where we have $k$-times the $1$.
[b]Induction Step:[/b] We'll prove $(a_{(k+1)n+1},a_{(k+1)n+2},\dots,a_{(k+2)n})=(0,0,\dots,0,1,\dots,1)$ with $k+1$-times the $1$. As for that, we'll induct again to show \[ a_{(k+1)n+1}=a_{(k+1)n+2}=\dots+a_{(k+1)n+n-k-1}=0. \] By $\sum_{i=1}^n a_{(k+1)n+i}=k+1$ we'd be done. Assume $a_{(k+1)n+1}=1$. Then \[ k+1=a_{kn+2}+a_{kn+3}+\dots+a_{(k+1)n+1} < a_{(k+1)n+2}+a_{(k+1)n+3}+\dots+a_{(k+2)n}+a_{(k+2)n+1} \leq k+1 \] by the Induction Hypothesis. Contradiction. Thus $a_{(k+1)n+1}=0$. So now assume \[ a_{(k+1)n+1}=a_{(k+1)n+2}=\dots=a_{(k+1)n+i} = 0 \quad \text{for all} \quad i \leq N \leq n-k-2. \] Again assume $a_{(k+1)n+i+1}=1$. Then \begin{align*} k+1 = a_{kn+i+2}+\dots+a_{(k+1)n+i+1} <k+2 &\leq a_{(k+1)n+i+2}+\dots+a_{(k+2)n+i+1}
\\ < k+3 &\leq a_{(k+2)n+i+2}+\dots+a_{(k+3)n+i+1}
\\ \dots
\\ < k+i+2 &\leq a_{(k+i+1)n+i+2}+\dots+a_{(k+i+2)n+i+1}. \end{align*} Here we've used that $a_{(k+1)n+i+2}+\dots+a_{(k+2)n}=k$ Thus $a_{(k+2)n+1}+\dots+a_{(k+2)n+i+1}=2$. With that $a_{(k+2)n+i+2}+\dots+a_{(k+3)n}=k$ and thus $a_{(k+3)n+1}+\dots+a_{(k+3)n+i+1}=3$ and so on. But the last line suggest \[ a_{(k+i+2)n+1}+a_{(k+i+2)n+2}+\dots+a_{(k+i+2)n+i+1} = i+2. \] But those are just $i+1$ terms. Contradiction. It's easy the verify that those indizes are well-defined for what we need. That ends the induction and thus also the other induction. It's easy to check that the claimed sequence indeed is a solution.

For $1\leq k\leq n^2-n+1$, let $b_k = a_k + a_{k+1} + \cdots + a_{k+n-1}$. It follows from the conditions given that $b_1<b_{n+1}<\cdots <b_{n^2+1}$, but each of these numbers is an integer from $0$ to $n$ inclusive, hence all these integers appear exactly once, and $b_1=0, b_{n^2+1} = n$. So the first $n$ terms of the sequence are $0$ and the last $n$ terms are $1$.

Let $S_i = \{ b_i, b_{i+n}, \cdots b_{i+n^2-n}\}$ for $2\leq i \leq n$. This is a strictly increasing sequence of $n$ integers from $0$ to $n$ inclusive. Let their sum be $T$. Also,
$a_1+a_2+\cdots +a_{n^2+n} = 0+1+\cdots +n= T + (n-i+1)$. Here we used the fact that the first $n$ terms are $0$ and the last $n$ terms are $1$. Thus the missing element in $S_i$ is precisely $n-i+1$. Now we have shown that each sum of $n$ consecutive terms is uniquely determined, so there's at most one such sequence.

It is obvious that $B_1B_2\cdots B_{n+1}$ works, where these are blocks of $n$ terms, and in $B_i$ everything is $0$ except for the last $i$ terms.

[hide=Solution]
Define $R(x)$ to be the remainder when $x$ is divided by $n$, and $S(x) = \lfloor \frac{x}{n} \rfloor+R(x)-n$. We claim that the only such sequence is the one defined as follows: $a_i = 1$ iff $S(x-1)$ is nonnegative.

Now, define $T(x)$ to be $a_{x+1}+a_{x+2}+...+a_{x+n}$ and let $f(x)$ denote the minimal amount $i$ such that $T(i) \geq x$. First, we note that $T(0), T(n), ..., T(n^2)$ give $n+1$ strictly increasing values between $0$ and $n$, so $T(0) = 0$. Furthermore, we know that $T(kn) = k$. Thus, $1 \leq f(1) \leq n$. Now, note that if $x>0$ and $f(x)=i$, then $a_{i+n}=1$. Otherwise, $T(i-1) = T(i)+a_i-a_{i+n} \geq T(i) = x$ and thus $f(x) < i$.

Lemma: If $f(x) = i$ and $x>0$, then $f(x+1) \leq i+n-1$
Proof: Note that $T(i+n)\geq T(i)+1$(by condition b) of the problem), so $T(i+n-1) = T(i+n)-a_i+a_{i+n} \geq 1-a_i+(n+1) \geq n+1$, completing the lemma.

Now, by Lemma 1, combined with the fact that $f(1) \leq n$, $f(i) \leq (n-1)i+1$, so $f(n) \leq (n^2-n+1)$ with equality iff $f(x) = (n-1)i+1 \forall i$. Furthermore, we must have equality, otherwise $T(f(n)+n) \leq n = T(f(n))$, which is a contradiction to condition b).

Thus, the first $1$ must be at $a_{2n}$ because $f(1) = 1$. Furthermore, since $T(kn) = k$, we can inductively show the first $1$ in the interval $[(k-1)n+1, kn]$, noting that there must be at $k(n-1)+2$ and there must be exactly $k-1$ ones in the interval, and thus $a_{kn-k+2}, a_{kn-k+3}, ..., a_{kn} = 1$ for all $k$, with base case $k=2$, which is true because $f(1) = n$.

Now, we prove that constructed sequence(which is the same as the original sequence described at the beginning of the solution) works. Note that $a_{k+n} \geq a_k \forall k$, so $T(k+1) = T(k) + a_{n+k+1} - a_{n+1}$, so $T(k+1) \geq T_k$. Thus $T$ is nondecreasing, and we know that $(n-1)i+1$ is the minimal value $x$ such that $T(x) = i$ from above. Thus, $T(x) =  \lfloor \frac{x-1}{n-1} \rfloor$(unless $x=n^2$, then $T(x) = n$, and clearly $\lfloor \frac{x+n}{n-1} \rfloor = \lfloor \frac{x+1}{n-1} \rfloor+1 \geq \lfloor \frac{x}{n-1} \rfloor+1 $ for all $x$, so it satisfies condition b).[/hide]

The only possible config is have for each $i$th block of $n$ terms from left consisting of $n-i$ zeros and then $i$ ones.

First, consider all blocks of $n$ $a_i$'s such that the first term $a_j$ of each block has $j \equiv 1 \pmod{n}$. There are $n+1$ such blocks and since they are disjoint, the blocks must have sums $0, 1, \cdots n$ respectively from left to right.

Then consider all blocks of $n$ such that first term $a_j$ of each block has $j \equiv 2 \pmod{n}$. There are $n$ such blocks and by the previous observation the sums of all terms in these blocks is $\frac{n(n+1)}{2}-(n-1)$. The sums on these blocks are distinct and so it must be that their sums are $0, 1, \cdots n-2, n$ respectively from left to right.

Then we can fill out $0$'s and $1$'s from right to left and we see that we are forced to have the said configuration.

There is a unique valid sequence, which is described below:
[asy]
	int n = 6;
	for (int i = 0; i < n; i += 1) {
		draw((n-1, -i)--(0, -i-1), gray(0.6));
	}
	for (int i = 0; i <= n; i += 1) {
		draw((0, -i)--(n-1, -i), gray(0.6));
	}
	for (int i = 0; i < n; i += 1) {
	for (int j = 0; j <= n; j += 1) {
		label(string(i + j >= n ? 1 : 0), (i, -j));
	}
	}
[/asy]
It is clear that this works.

Now we show it is the only one. Write the sequence as an $(n+1) \times n$ matrix as above; then by condition the row sums are $0, \dots, n$ in order. In particular the top row is all-zero while the bottom row is all-one.

Define the partial sums $s_k = a_1 + \dots + a_k$, and let $i \in \{1, \dots, n-1\}$ be an integer. Then the $n$ nonnegative integers
\[s_{i+n} - s_i, s_{i+2n} - s_{i+n}, \dots, s_{i+n^2} - s_{i+n^2-n}\]
form a strictly increasing sequence and sum to
\[s_{i+n^2} - s_i = (0 + 1 + \dots + n) - (n - i)\]
and so must be exactly $\{0, 1, \dots, n\} \setminus \{n-i\}$ in order.

In particular each $s_{k+n} - s_k$ is forced, so in light of $a_1 = \dots = a_n = 0$ the entire sequence is forced.

It seriously felt like an algebra problem.
[quote=orl]Let $ n > 1$ be an integer. Find all sequences $ a_1, a_2, \ldots a_{n^2 \plus{} n}$ satisfying the following conditions:
\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 \plus{} n;
\]
\[ \text{ (b) } a_{i \plus{} 1} \plus{} a_{i \plus{} 2} \plus{} \ldots \plus{} a_{i \plus{} n} < a_{i \plus{} n \plus{} 1} \plus{} a_{i \plus{} n \plus{} 2} \plus{} \ldots \plus{} a_{i \plus{} 2n} \text{ for all } 0 \leq i \leq n^2 \minus{} n.
\]
[i]Author: Dusan Dukic, Serbia[/i][/quote]
We claim that the only sequence possible is the following:
$$\underbrace{0,0,\cdots 0}_{n}\underbrace{0,0, \cdots,0,1 }_{n}\underbrace{0,0, \cdots,0,1,1 }_{n} \cdots \underbrace{1,1, \cdots,1,1 }_{n}$$
It is not hard to see that this works. Now we show that this is the only possibility.

Let $\mathcal{T}(k)$ denote the number of $1$s in the block $[a_{k}, a_{k+n+1}].$ Then the condition translates to $\mathcal{T}(k)<\mathcal{T}(k+n)$ for all $0 \le k \le n^2-n.$ For simplitcity, define $f$ by $f_i(k)=\mathcal{T}(ni+k).$ For instance, when $n=3;$
$$0, \underbrace{0,0,0}_{f_2(0)=0}\underbrace{0,1,0}_{f_2(1)=1}\underbrace{1,1,1}_{f_2(2)=3},1,1$$

We will now show that each $\mathcal{T}(k)$ has a unique value. Since the solution given before works, hence it would be the only solution.

Clearly $\mathcal{T}(ni+1)=i$ for all $1 \le i \le n.$ In particular the first $n$ elements are zeros while the last $n$ elements are $1$s. Thus by comparing terms we see
\begin{align*}
f_2(0)+f_2(1)+\cdots +f_2(n-1) &=f_1(0)+f_1(1)+\cdots f_1(n)-0- \underbrace{\left(1+1+\cdots+1\right)}_{n-1}  \\
f_2(0)+f_2(1)+\cdots +f_2(n-1) &=0+1+\cdots+(n-1)+1
\end{align*}
Noting that $f_2(i)<f_2(j)$ for all $i<j$ gives a unique value designation to each of $f_2(i).$ In fact, this is $f_2(i)=i$ for all $0 \le i \le n-2$ and $f_2(n-1)=n.$

We can further repeat this procedure to assign values to each of $f_3(i), f_4(i), \cdots.$ and hence conclude the result. $\square$

[b]EDIT:[/b] I realized that this is just a detailed form of the above solution.

First, note that by the given inequality, we have $$a_1+a_2+\dots+a_{n}<a_{n+1}+a_{n+2}+\dots+a_{2n}<\dots<a_{n^2+1}+a_{n^2+2}+\dots+a_{n^2+n}.$$ Since there are only $n+1$ possible values of the sum of any $n$ consecutive terms of the sequence, this implies that $$a_1+a_2+\dots+a_{n}=0,$$$$a_{n+1}+a_{n+2}+\dots+a_{2n}=1,$$$$\vdots$$$$a_{n^2+1}+a_{n^2+2}+\dots+a_{n^2+n}=n.$$ In particular, the first $n$ terms must be $0$ and the last $n$ terms must be $1.$\\\\ We claim that for $i\in\{1,2,\dots,n\},$ we have $$a_{i}=a_{n+i}=\dots=a_{(n-i)n+i}=0,$$$$a_{(n-i+1)n+i}=a_{(n-i+2)n+i}=\dots=a_{n^2+i}=1.$$ To show this, we induct backwards on $i.$ \\\\ For the base case $i=n,$ let $x_k=a_{kn+1}+a_{kn+2}+\dots+a_{kn+n-1}$ for $k\in\{1,2,\dots,n-1\}.$ Using our work before and the given inequality, we have $$x_1<1-x_1+x_2<2-x_2+x_3<\dots<n-x_{n-1}+n.$$ Since $x_k\in\{k-1,k\}$ for all $k,$ the only solution to this inequality is $x_k=k-1$ for all $k,$ implying the claim.\\\\ Now suppose the claim is true for $i=n,n-1,\dots,m,$ and let $x_k=a_{kn+1}+a_{kn+2}+\dots+a_{kn+m-2}$ for $k\in\{1,2,\dots,n-1\}.$ By the given inequality, $$x_1<2-x_1+x_2<2-x_2+x_3<\dots<n-x_{n-1}+n.$$ But by the inductive hypothesis, we know $x_k=0$ if $k\le n-m+1$ and $x_k\in\{k-n+m-2,k-n+m-1\}$ if $k>n-m+1.$\\\\ Therefore, the only solution is $x_1=x_2=\dots=x_{n-m+1}=0$ and $x_{k}=k-n+m-2$ for $k>n+m-1,$ implying the claim.\\\\ We are done by induction.

It will be helpful to define the following notation: Let the 1-block be the $n$ numbers from $a_1$ to $a_n$, define 2-blocks, etc analogously. Additionally, let $S(m, x)$ be the sum of the numbers from indices $m$ to $x$. Notice that \begin{align*}
    S(1, n) < S(n + 1, 2n) < \cdots < S(n(n - 1) + 1, n^2) < S(n^2 + 1, n^2 + n).
\end{align*} There are $n + 1$ blocks total, and the only possible sums are $0$ to $n$. Thus, 1-block has sum 0, 2-block has sum 1, and so forth.

Now, we claim that the only possible sequence is when the ones in each block are pushed as far to the right as possible; for example, the two $1$s in the 3-block would be pushed to $3n$ and $3n - 1$.

For our base case, assume that the $1$ in the $2$-block appears at index $n + j$, where $j < n$ for contradiction. Then, draw new boxes from $j + 1$ to $n + j$, and so forth. Evidently, \begin{align*}
    1 = S(j + 1, n + j) < S(n + j + 1, 2n + j) < \cdots < S(n(n - 1) + j + 1, n^2 + j),
\end{align*} which forces there to be $1, 2, \cdots, n$ ones in each of the new boxes. Observe now that $S(n(n - 1) + j + 1, n^2)$ is all ones for a total of $n - j$ ones (this is because the n+1-block is all ones). By similar logic, we need $S(n(n - 1) + 1, n(n - 1) + j) = j - 1$, but then this forces $S(n(n - 2) + j + 1, n(n - 1)) = n - j$ and so forth which is a contradiction.

The inductive step is near identical to the base case. The only difference is that we note that the $k + 1$th block has its earliest $1$ at $n - k$, and as a result the boxes we draw in the inductive step will contain all the pushed-forward ones in the $k$th block.

That was all a little dense, so here's an illustrative example. Let's take a look at the sequence $0000, 00 | 01, 01| |--, --| |--, 11|11$. The dashes represent the new boxes we draw in the inductive step, call them Crates 1, 2, and 3. Crate 1 has a sum of $2$, and because there are three crates total, we know from the givens that Crate 2 must have a sum of 3, and Crate 3 has a sum of 4.

As a result, our new sequence is $0000, 00 | 01, 01| |--, (--)| |11, 11|11$. However, because we know that box 4 must contain $3$ ones total, we know that the parenthesed area can only contain one $1$ at maximum. As a result, we are forced to draw the following: $0000, 00 | 01, 01| |11, --| |11, 11|11$, which is a contradiction since Box 3 here has a sum of $3$ which is too much.

It is clear that $a_1+a_2+\cdots+a_n<a_{n+1}+\cdots+a_{2n}<\cdots<a_{n^2+1}+\cdots+a_{n^2+n}$. There are $n+1$ sums, so the first is $0$, the next is $1$, etc. In particular, $a_1=a_2=\cdots=a_n=0$ and $a_{n^2+1}=a_{n^2+2}=\cdots = a_{n^2+n}=1$.

We claim that $a_{n+n}$ is the sole nonzero $a_{n+i}$ among $1\le i\le n$. This is because otherwise considering the sums $a_{kn+i+1}+\cdots+a_{kn+n+i}$ for $0\le k\le n-1$ implies that each successive sum contains all nonzero elements of $a_{kn+n+1},a_{kn+n+2},\dots,a_{kn+2n}$. A similar argument implies that only $a_{3n}$ and $a_{3n-1}$ of $a_{2n+1},\dots,a_{3n}$ are nonzero. Continuing on this way, we can uniquely characterize the sequence as the one for which each $a_{kn+i}$ with $1\le i\le n$ is nonzero iff $n+1-k\le i$. It is not hard to check this sequence works, so done.

I claim that there is one unique solution.

We proceed with induction on $n$,the base case is trivial and we obviously have one unique solution ($a_1<a_2$).

Now the inductive step:

Write $s_i = a_{i+1}+\ldots a_{i+n}$. Then, note that we have
\[0\leq s_0 < s_n < \cdots s_{(n-1)\cdot n} < s_{n^2} \leq n\]
Thus, since these values are integers, and we have $n+1$ choices and $n+1$ values, we have $s_{nk} = k$ for all $k$.

Now, let $i$ be the smallest $i$ such that $a_i=1$. Note that we are guaranteed that $n+1\leq i \leq 2n$. Thus,
\[1\leq s_{i-n}< s_i < s_{i+n}\ldots s_{i+(n-2)n}\leq n\]
There are $n$ choices, and $n$ values, so we have $s_{i+nr} = r+2$. Importantly,
\[\sum_{k=0}^{n} s_{nk} = \sum_{r=-1}^{n-2} s_{i+nr}\]
Thus, since the LHS represents all 1s, all 1s must also be in the $i+nr$ ranges, so there cannot be any 1s in between $i+(n-2)n$ and $n^2+n$, but this is absurd because $s_{n^2}=n$ guarantees that for all $n^2+1\leq x\leq n^2+n$, $a_x=1$. Thus, we must have that \[i+(n-2)n=n^2 \Longrightarrow i=2n\]
By considering $s_{nk-1}$
\[2\leq s_{2n-1} < s_{3n-1} < \ldots < s_{n^2-1} \leq n\]
we get that $s_{nk-1}=k$. Thus, we must have that $a_{3n}=1$ and so on so that $a_{nk}=1$ for all $k\geq 2$. Thus, we can now remove $(1,2,\ldots n)$ and $(2n,3n,\ldots n^2+n)$, at which point we have reduced to the $n-1$ case because we've removed exactly 1 from each series of $n$. Thus, by the inductive hypothesis there is exactly one solution.

We can construct such a construction by taking a pattern of the form $a_{kn-r}=1$ for $0\leq r<n$ if and only if $r\leq k-2$.

[hide=My Solution]Let $b_i=\sum_{j=1}^{n}a_{i+j}.$ Therefore, $b_0<b_n$ from the second condition. We also have $b_n<b_{2n}$ and so on. This implies that the sequence $b_0,b_n,\dots,b_{n^2}$ is increasing. However, the lower bound of $b_i$ is $0$ while the upper bound of $b_i$ is $1\cdot n=n$, and since all $b_i$ are integers, we must have that $b_{kn}=k$ where $k$ is a nonnegative integer. However, from the definition of $b_i$, this means that $a_1=a_2=\dots=a_n=0.$ We also know that $\sum_{j=1}^{n^2+n}a_j=\sum_{m=0}^{n}b_{mn}=\frac{n(n+1)}{2}.$ This means that for all positive integers $y<n$, we know that $$b_{y}+b_{n+y}+\dots+b_{n^2-n+y}=\sum_{j=1}^{n^2+n}[a_j]-\sum_{k=1}^{y}[a_k]-\sum_{l=n^2+y+1}^{n^2+n}[a_l]$$$$=\frac{n(n+1)}{2}-(n-y)=\frac{n(n+1)}{2}-n+y$$ because $b_{n^2}=n.$ Because we have that $b_y<b_{n+y}<\dots<b_{n^2-n+y}$ and they are all integers, the lower bound is $0$, and the upper bound is $n$, and since there are exactly $n$ terms, the missing term must be $n-y.$ This means that $b_{kn+y}=k$ if $k<n-y$ and $k+1$ if $k\ge n-y.$ This means that, for any nonnegative integer $k<n$, we know that $b_{kn+y}=k$ if $y<n-k$, even with $y=0$ because $k<n$. We also know that $b_{kn+y}=k+1$ if $y\ge n-k$, even with $y=n$ because $b_{(k+1)n}=k+1.$ This means that $b_i$ increases if and only if it is from $b_{(k+1)(n-1)}$ to $b_{(k+1)(n-1)+1}$ for nonnegative $k<n$, and also, $b_{k(n-1)+1}=b_{k(n-1)+2}=\dots=b_{(k+1)(n-1)}.$ This means that $b_i=\lfloor\frac{i-1}{n-1}\rfloor$ except for $i=n^2$, in that case, $b_{n^2}=n.$ Therefore, we know that $b_{i+1}-b_i=a_{i+n+1}-a_{i+1}$ is $1$ for $i=k(n-1)$ with $k\le n$ and $0$ for everything else. Thus, for positive integer $z$ with $1\le z\le n$ and positive integer $w$,
$$\boxed{a_{wn+z} = 0 \;\;\text{if $w\le n-z$ and}\;\; 1\;\; \text{if $w>n-z$}}.$$
As we have constructed and proven all $a_i$, we are done.[/hide]

Break the sequence into $n+1$ groups of $n$ numbers each. Using the second condition for multiples of $n$, we get that the amount of $1$s in a group increases with the position of the group. As there are only $n+1$ possible values for the number of $1$s in a group, the $i$th group will have $i-1$ ones.

We claim that the answer consists of the $i$th group consisting of $i-1$ ones at the end of the group, and zeros at the beginning where needed. This can be shown to work. We will prove this using induction. It is easy to show this for $n=2$.

Suppose that the claim is true for $n=k-1$. We will prove the statement for $n=k$.

[b][color=#f00]Claim:[/color][/b] For any $n$, the last number of every group other than the first must be a $1$.
[i]Proof.[/i] Suppose that the last number in the $x$th group is $0$, where $x>1$. Plugging in $(n-1)x-1$ into the second condition gives us that the sum of the last number of the $x-1$th group and the first $n-1$ numbers of the $x$th group is less than the sum of the last number of the $x$th group and the first $n-1$ numbers of the $x+1$st group. However, the former is at least $x-1$ and the latter is at most $x$, so equality must hold, meaning all the ones in the $x+1$st group are in the first $n-1$ numbers. This means that the last number in the $x+1$th group is $0$.

Continuing this reasoning, we will eventually get that the last number of the last group, or $a_{n^2+n}$, is $0$, which is impossible. $\blacksquare$

Consider the sequence consisting of the $k^2-k$ numbers $a_{ln+m}$ for $1\le l\le k, 1\le m\le k-1$ in order. We claim that this sequence satisfies both conditions for $n=k-1$. Indeed, the first condition is obviously satisfied. The second condition is equivalent to the second condition for the $n=k$ sequence because we simply take out the last number out of a group (for the $n=k$ sequence) for each side, which is $1$ by the claim.

Therefore, by the induction hypothesis, there is only one possibility for $a_{ln+m}$ for $1\le l\le k, 1\le m\le k-1$. All other values are already determined, and combining these with the fixed values for $a_{ln+m}$ for $1\le l\le k, 1\le m\le k-1$, we get the answer stated above. The induction is complete, and we are done.

Let $b_i=a_i+a_{i+1}+\dots+a_{i+n}$. Trivially, we obtain $b_{kn+1}=k$, due to the second condition. It follows that $b_{n^2+1}=n$ or that $a_i=1$ for all $n^2+1\le i\le n^2+n$. Now, we have that
\[\sum_{k=0}^nb_{kn+1}=\sum_{k=1}^{n^2+n}a_k=\frac{n(n+1)}{2}=s_1\]
\[\sum_{k=0}^{n-1}b_{kn+2}=\sum_{k=2}^{n^2+1}a_k=s_1-(n-1)=s_2\]
\[\vdots\]
\[\sum_{k=0}^{n-1}b_{kn+n}=\sum_{k=n}^{n^2+n-1}a_k=s_1-1=s_n.\]
Now, consider the sequence $b_i, b_{n+i}, \dots, b_{n^2-n+i}$ for some $1<i\le n$. It's easy to see that this sequence contains every number from $0$ to $n$, inclusive, except for one. From our sums earlier, we obtain that this number is actually $n+1-i$.

Now, I claim that the sequence $b_{kn+1}$ to $b_{kn+n}$ contains $k$ copies of $k+1$ and $n-k$ copies of $k$, in sorted order. We show this by induction. The base case ($k=0$) is trivial. Now, suppose that $k-1$ satisfies the given. Then clearly we have $b_{kn+i}=b_{kn+i-n}+1$ for all $1\le i\le n$ except for $i=n-k+1$, in which case we obtain $b_{kn+n-k+1}=b_{kn-k+1}+2$. This implies the result.

Finally, I claim that $\boxed{\text{the sequence }a_{kn+1}\text{ to }a_{kn+n}\text{ contains }n-k\text{ copies of }0\text{ and }k\text{ copies of }1\text{ in sorted order}}$. We show this by induction; the base case is trivial. Consider $b_{kn+1-k}$ and the sequence $a_{kn+2-k}$ to $a_{kn}$. Clearly this sequence contains all of the $1$s that contribute to the sum in $b_{kn+1-k}$. It follows that all values from $a_{kn+1}$ to $a_{kn+n-k}$ are equal to $0$. However, $b_{kn+1}$ is equal to $k$, which implies that the $k$ elements from $a_{kn+n-k+1}$ to $a_{kn+n}$ are equal to $1$. Therefore, we are done. $\blacksquare$

Let $s_i=a_{i + 1} + a_{i + 2} + \ldots + a_{i + n},0\le s_n\le n.$ Note that $s_0<s_n<\dots<s_{n^2}$ so $s_{kn}=k.$ In particular, $a_1=a_2=\dots=a_n=0,a_{n^2+1}=a_{n^2+1}=\dots=a_{n^2+n}=1$ Let $1\le k\le n-1$, then $s_k< s_{k+n}<\dots < s_{k+n^2-n}$ and $s_k+s_{k+n}+\dots+s_{k+n^2-n}=\frac{n(n+1)}{2}-(n-k)$ which means that this sequence contains every integer from $1$ to $n$ excluding $n-k.$ Thus, from $a_1=a_2=\dots=a_n=0,a_{n^2+1}=a_{n^2+1}=\dots=a_{n^2+n}=1$ we get the same construction everyone else has posted.

We claim that the in the sequence, the first $n$ terms are all 0, in the next $n$, only the last term is 1, in the next $n$, only the last 2 terms are 1, and so on, until in the last $n$, all terms are 1. This sequence clearly works.\\

Note that each block of $n$ must contain strictly more 1's than the previous block. This implies that there must be no 1's in the first $n$, one 1 in the next $n$, two 1s in the next $n$, and so on until $n$ ones in the last $n$.\\

Consider the prefix sums of this sequence. We will put the prefix sums in $n+1$ rows of $n$ each so that when the grid is read by rows, it gives the prefix sum sequence. Note that by our earlier observation, the last column must contain $0,1,3,6\cdots n(n+1)/2.$ Additionally, condition (b) is equivalent to each column forming a strictly convex sequence. Now, consider the second to last column of these prefix sums. This column must end in $n(n+1)/2 -1$ since the final $n$ terms are all 1's, so the final row of this grid contains consecutive integers. However, note that the minimal convex sequence starting with $0,1$ with a length of $n+1$ ends with $n(n+1)/2$, contradiction, so the second to last column must start with $0,0$ instead. Note that consecutive prefix sums can differ by no more than 1, so the second to last column must  be $0,0,2,5,9,14\cdots n(n+1)/2 -1.$ We can then repeat this argument with previous columns. For example, in the third to last column, it must start with $0,0,1$ since the minimal convex sequence starting in $0,0,2$ reaches $n(n+1)/2-1$ on term $n+1$, but it needs to end in $n(n+1)/2-2,$ so it must start in $0,0,1$, and we can also use the argument that it can never fall behind the second to last column by more than 1. Repeating this argument with all previous columns, we get a unique possible grid of prefix sums, from which we can uniquely recover the original sequence, and since the sequence we showed earlier works, we are done.

Odd problem.

Divide the sequence into subsequences $s_1 = (a_1, a_2, \dots, a_n)$, $s_2 = (a_{n+1}, a_{n+2}, \dots, a_{2n})$, and so on. The only such sequences are those with $s_i = (0, 0, 0, \dots, 1, 1, \dots, 1)$, where there are $i-1$ ones consecutively.

To prove this, notice that each subsequence $s_i$ contains a strictly greater number of ones than any previous subsequence; by Pigeonhole it follows that it must contain exactly $i-1$ ones. Now, suppose for the sake of contradiction that there exists a minimal index $k$ for which $s_k$ contains a one before index $n-k+2$ in that subsequence. Call this index $i$.

By minimality, we know that the $n$ digits leading up to index $i$ (spanning two subsequences) contains at least $k-1$ ones because due to the condition $i \leq n-k+1$, all zeroes at the tail of the previous subsequence are contained in these $n$ digits. This means that there are at least $k$ ones in the $n$ digits immediately following; in other words, all $k$ ones in the subsequence $s_{k+1}$ must fall in these $n$ digits, and thus strictly before the index $n-k+2$ in that sequence.

This implies that the next list $s_{k+1}$ also satisfies the aforementioned condition, so inductively, the list $s_{n+1}$ must also satisfy this condition. But this is absurd, because clearly there are no elements of $s_{n+1}$ that are before the element indexed $n-(n+1) + 2 = 1$.

We claim that there is one unique sequence. First, we will construct this sequence, which will be made up of $n+1$ blocks of $n$ digits. The first block will be of $n$ zeros, the next will be of $n-1$ zeros with the last digit being $1$, and the second block will have the last two digits as $1$, and so on and so forth. E.g. for $n=3$, we have the sequence $000001011111$.

Clearly this construction works, as for each time we "increase" or move to the next block of $n$ numbers from any block of $n$ numbers(not necessarily the ones specified in the previous definition, a "block" could mean index $2$ to index $n+1$), we "exchange" a zero for a $1$, adding one to the sum.

[s]Now we must prove that there are no other sequences. Notice that if we section off the $n^2+n$ numbers into $n+1$ disjoint blocks of $n$ (there is only one way to do this), we have that the sum of the first block must be $0$, the next is $1$, and so on and so forth, since these sums must be increasing. Using this and filling out the sequence from left to right, we find that we must have the configuration, and we are done. [/s]

[color=#9a00ff]FS found by [b]bobthegod78[/b]

Now we must prove that there are no other sequences. Let the variable $S_{m,m+k}$ for $k>0$ denote the sum $a_m+a_{m+1}+a_{m+2}+\dots{}+a_{m+k}$. Notice that
\[S_{1,n}<S_{n+1,2n}<S_{2n+1,3n}<\dots{}S_{n^2+1,n^2+n}\]
which means that $S_{cn+1,cn+n}=c$ for any integer $c$ from $0$ to $n$ inclusive.

Now we will prove that $a_{n+1}=a_{n+2}=\dots{}=a_{2n-1}=0$, and $a_{2n}=1$. Notice that since $S_{n+1,2n}=1$, we have that exactly one of the $n$ values between $a_{n+1}$ and $a_{2n}$ is $1$, and because $S_{1,n}=0$, we also must have that $a_1=a_2=\dots{}=a_n=0$.

FTSOC, assume that $a_{n+1}=1$. From the condition, we have that
\[S_{2,n+1}<S_{n+2,2n+1}<\dots{}S_{n^2-n+2,n^2+1}\]
and since $S_{2,n+1}=1$, we must have that $S_{n+2,2n+1}=2$. However, this implies that $a_{2n+1}=2$, a clear contradiction.

Similarly, if we let $a_{n+c}=1$ for some $0<c<n$, we will find that it implies that $a_{cn+1}=2$, a contradiction. Therefore, we must have that $a_{2n}=1$. Similarly, we can do this for the other $n-1$ blocks of binary with sums $2$, $3$, $\dots$, $n-1$ by focusing on the first $1$ that appears in that block, and we can conclude that it will always be $a_{cn-c+2}=a_{cn-c+3}=\dots{}=a_{cn}=1$. Therefore there is only one sequence, and we are done. [/color]

[hide=Solution]
There is only one such sequence $\{a_k\}$ -- it's the one with $(a_1, a_2, \cdots, a_n) = (0, 0, \cdots, 0)$, $(a_{n+1}, a_{n+2}, \cdots, a_{2n}) = (0, 0, \cdots, 0, 1)$, $(a_{2n+1}, a_{2n+2}, \cdots, a_{3n}) = (0, 0, \cdots, 0, 1, 1)$, and so on, $(a_{n^2-n+1}, a_{n^2-n+2}, \cdots, a_{n^2}) = (0, 1, \cdots, 1, 1)$, and $(a_{n^2+1}, a_{n^2+2}, \cdots, a_{n^2+n}) = (1, 1, \cdots, 1, 1)$. This works, since clearly this satisfies (a). For (b), note that $a_i \leq a_{i+n}$, since $a_r, a_{r+n}, a_{r+2n}, \cdots$ is all zeroes until one point, and then it switches to all ones. Thus, for any $0\leq i\leq n^2-n$, $a_{i+1} + a_{i+2} + \cdots + a_{i+n} \leq a_{i+n+1} + a_{i+n+2} + \cdots + a_{i+2n}$. Now, we wish to show that equality cannot be held. If equality held, then equality must have held everyhwere in our original estimates, i.e. $a_{i+1} = a_{i+n+1}$, $a_{i+2} = a_{i+n+2}$, $\cdots$, $a_{i+n} = a_{i+2n}$. Now, write $i + 1 = nk+r$ for some nonnegative integers $k, r$ with $1\leq r\leq n$ Since $i\leq n^2-n$, $k\leq n-1$. Then, $a_{nk+r}, a_{nk+(r+1)}, \cdots, a_{nk+n}$ are the same as $a_{nk+r+n}, a_{nk+(r+1)+n}, \cdots, a_{nk+2n}$, respectively. Thus, since $a_{nk + (n-k)} = 0$ and $a_{nk+(2n-k)} = 1$, we must have $r > n-k$ (consequently, $r > 1$). Then, $nk+r < n(k+1) + (n-k-1) < n(k+1)+r$, so $a_{n(k+1) + (n-k-1)}$ and $a_{n(k+1) + (2n-k-1)}$ are the same. However, $a_{n(k+1) + (n-k-1)} = 0$ and $a_{n(k+1) + (2n-k-1)} = 1$, a contradiction. Thus, equality cannot hold everywhere and so (b) is satisfied.

Now, we show that $\{a_i\}$ must be of this form. Firstly, note that $$0\leq a_1 + a_2 + \cdots + a_n < a_{n+1} + a_{n+2} + \cdots + a_{2n} < \cdots < a_{n^2+1} + a_{n^2+2} + \cdots + a_{n^2+n} \leq n.$$ Thus, since the $a_i$ are integers, it follows that $a_1 + a_2 + \cdots + a_n = 0$, $a_{n+1} + a_{n+2} + \cdots + a_{2n} = 1$, $\cdots$, and $a_{n^2+1} + a_{n^2+2} + \cdots + a_{n^2+n} = n$. Also, this implies that $a_1 = a_2 = \cdots = a_n = 0$,  $a_{n^2+1} = a_{n^2+2} = \cdots =a_{n^2+n} = 1$, and that $$a_1 + a_2 + \cdots + a_{n^2+n} = 0 + 1 + \cdots + n = \frac{n(n+1)}{2}.$$

We will now prove by strong induction that for any $k\geq 0$, we have that $a_{nk+r} = 0$ for $r\leq n-k$ and $a_{nk+r} = 1$ for $r > n-k$. We already have established the base case $k = 0$ since we have proven that $a_1 = a_2 = \cdots = a_n = 0$.

Now, assume that for all $0\leq i\leq k$, the result is true for $i$. We will show that the result is true for $k+1$.

Assume the contrary. Then, there exists a $1\leq r \leq n-(k+1)$ for which $a_{n(k+1)+r} = 1$. Suppose $r$ was the smallest such $1\leq r\leq n-(k+1)$ for which $a_{n(k+1)+r} = 1$. Then, $a_{n(k+1)+s}$ for $1\leq s < r$ are all $0$, and $a_{nk+t}$ for $r+1\leq n-k < t \leq n$ are all $1$ (if there exists such $t$), and $0$ otherwise. Thus, $$a_{nk+r+1} + a_{nk+r+2} + \cdots + a_{nk+r+n} = k+1.$$ Then, we have that $$k+1\leq a_{nk+r+1} + a_{nk+r+2} + \cdots + a_{nk+r+n} < a_{n(k+1) + r + 1} + a_{n(k+1) + r+2} + \cdots + a_{n(k+1) + r + n} < \cdots < a_{n(n-1) + r + 1} + a_{n(n-1) + r + 2} + \cdots + a_{n(n-1) + r + n} \leq n,$$ so since the $a_i$ are integers, it follows that $a_{n(k+1)+r+1} + a_{n(k+1)+r+2} + \cdots + a_{n(k+1) + r + n} = k+2$, $\cdots$, $a_{n(n-1) + r + 1} + a_{n(n-1)+r+2} + \cdots + a_{n(n-1)+r+n} = n$. Now, note that the sum of all the $a_i$ for $i \leq nk+r$ is $1 + 2 + \cdots + (k-1)$, since $a_{nk+1}, a_{nk+2}, \cdots, a_{nk+r}$ are all zero, and $a_1+a_2+\cdots + a_n = 0$, $a_{n+1}+a_{n+2} + \cdots + a_{2n} = 1$, $\cdots$, $a_{n(k-1)+1} + a_{n(k-1)+2} + \cdots + a_{n(k-1)+n} = k-1$. Now, $a_{n^2+r+1}, a_{n^2+r+2}$, $\cdots$, $a_{n^2+n}$ are all $1$, so their sum is $n-r$. Then, we have that
\begin{align*}
a_1 + a_2 + \cdots + a_{n^2+n} &= (1 + 2 + \cdots + (k-1)) + ((k+1) + (k+2) + \cdots + n) + (n-r) \\ &\geq (1 + 2 + \cdots + (k-1)) + ((k+1) + (k+2) + \cdots + n) + (k+1) \\ &> \frac{n(n+1)}{2} + 1,
\end{align*}
absurd since $a_1 + a_2 + \cdots + a_{n^2+n} = \frac{n(n+1)}{2}$ as established before. Thus, $\{a_i\}$ must be of this form, so we have proven that $\{a_i\}$ must be of this form and such $\{a_i\}$ works, so we're done. $\blacksquare$

We compare the $n+1$ groups of $n$ integers from beginning to the end. We can see that since there are $n+1$ possible values of the sum, the $n$th group must sum to $n-1$. We claim that the only configuration is when those $n-1$ $1$'s are at the end of that group. Specifically, for $n = 4$ our configuration is: \[00000001001101111111\]

We can see that this works since there are always increments for every increment in $n$

We claim by induction that we cannot have $a_x = 1$ where $x = y \cdot n + z$ where $y$ and $z$ are integers with $0 \leq y \leq n$ and $0 \leq z \leq n-y$.

Our base case is trivial.

Now for the inductive step. If we have $x = 1$, then we can compare the $n$ groups of $n$ numbers from \[ \{a_{z+1} \ldots a_{n+z} \}, \{a_{n+z+1} \ldots a_{2n+z} \} , \ldots \{a_{n^2 - n +z+1} \ldots a_{n^2 + z} \} \]. We can see from our inductive step that for each of first $y-1$ terms they equal every integer from $0$ to $y-2$ we can see that the term that contains $x$ is $y$ (also from our inductive step). Since there are $n$ terms in total, we can see that the next $n - y$ terms must occupy every integer from $y+1$ to $n$.

We can thus compute that the total number of $1$'s would be $\frac{(y-1)(y-2)}{2} + \frac{(n+y)(n-y+1)}{2}$. Simplifying this we can see that this is equal to $\binom{n}{2} + 1$. However, we know that the total number of $1$'s is $\binom{n}{2}$. Contradiction. $\blacksquare$

holy this took me 3 hours, imo should be placed in c3 lol
The key is to compare partial sums $s_i=a_{i+1}+\dots+a_{i+n}$. The condition implies $s_0<s_n<\dots<s_{n^2}$; in particular, since these values range from 0 to n, and there are n+1 groups, each group corresponds s.t. $s_{kn}=k$ (rigid!). We proceed by induction to show that there is one unique sequence, namely the sequence starting with the first n numbers 0, then the next string of n numbers having n-1 0s, with the last number being a 1, etc. by increasing the number of 1s that end in a string with length n. It's easy to check that this works. The base case n=1 is easily verified because the condition only needs to hold for i=0, meaning $a_1<a_2$ so it's just $0,1$ unique.

    Then take the smallest i s.t. $a_i=1$ (rigid!), along with $s_n=1$ implying $n+1\le i\le 2n$. Also, noting that the value of i makes $s_{i-n}$ contain $a_i$, meaning it's at least 1, we get $$1\le s_{i-n}< s_i <\dots s_{i+(n-2)n}\le n\implies s_{i+nj} = j+2\implies\sum_{k=0}^{n} s_{nk} = \sum_{j=-1}^{n-2}s_{i+nj}$$ since there are n choices and n values for $s_{i+nj}$; in particular, there cannot be any 1s in $(i+(n-2)n,n^2+n]$ (since otherwise the sums are nonequivalent), which implies, along with the fact that $s_{n^2}=n\implies a_x=1\forall n^2+1\le x\le n^2+n$, that we must have $i+(n-2)n=n^2\implies i=2n$ (otherwise there is a 1 in between).

    On the other hand, note that $s_{2n-1}$ contains $a_{2n}$ and another 1, since it has at least as many 1s as $s_{2n}=2$ minus 1 (if $a_{2n}$ were to equal 1). So we have the bound $$2\le s_{2n-1} < s_{3n-1} < \ldots < s_{n^2-1} \le n\implies 2=s_{2n-1}=s_{2n}\stackrel{a_{2n}=1}{\implies}a_{3n}=1;$$ in the same manner inductively (next we would take $3=s_{3n-1}=s_{3n}\stackrel{a_{3n}=1}{\implies}a_{4n}=1$), we can get that $a_{ln}=1$. Now, we can erase $a_x\forall x\in\{1,2,...,n,2n,3n,...,n^2+n\}$, which doesn't effect the condition, so we are done, because this has reduced the problem by 2n, which is indeed $n^2+n-(n-1)^2-(n-1)=2n$ into the hypothesis $(n-1)^2$. We already know that the terms that we erased did satisfy our claimed sequence, whence nothing changes. $\blacksquare$

edit: in OTIS version it said $n\ge 1$, which is why i used base case n=1; probably n=2 shouldn't be too nontrivial though even though its SIX VARIABLES.

also is it fine in contest if you extend domain given that there's no problems? im not sure if this is plausible someone pls answer

Kinda involved bruh.
Heres my sol.

 We claim there exists only one such sequence which is the sequence,
\begin{align*}
    a_1=a_2=\dots=a_n &= 0\\
    a_{n+1} = a_{n+2} = \dots = a_{2n-1}&=0 \ a_{2n}=1\\
    a_{2n+1} = a_{2n+2} = \dots = a_{3n-2}&=0 \ a_{3n-1}=a_{3n}=1\\
    &\vdots\\
    a_{n^2-n+1} =0 \ a_{n^2-n +2} = \dots = a_{n^2-1} = a_{n^2}&=1\\
    a_{n^2+1}=a_{n^2+2} = \dots = a_{n^2+n} &=1
\end{align*}
It is clear that this sequence clearly satisfies the required condition since if $a_k,a_{k+1},\dots,a_{l} \in \{a_{i+1},\dots,a_{i+n}\}$ are 1s for $0\leq i \leq n^2-n$, then so are $a_{k+n},a_{k+1+n},\dots,a_{l+n}$ an in addition so is $a_{k+n-1}$ confirming the strict inequality. We shall then show that this is the only sequence which satisfies the given conditions.\\

We have the following key claim.
[color=#f00][b]Claim :    [/b][/color] Consider the range $a_{(k-1)n+1},\dots,a_{kn}$. There exists no terms $a_i=1$ such that $i<kn-k+2$.

[i]Proof :  [/i]   Consider the above range. By way of contradiction, assume that there exists some $i<kn-k+2$ such that $a_i=1$. Let $m$ be the minimum such index. Let $m=kn-n+p$ ($p<n-k+2$). \\

    First, we have the chain of inequalities
    \[a_1+\dots+a_n < a_{n+1}+\dots + a_{2n} < \dots < a_{n^2}+\dots + a_{n^2+n}\]
    This means that since we only have 0s and 1s as terms, $a_(k-1)n+1+ \dots a_{kn} = k-1$ for all $1 \leq k \leq n+1$.
    First, note that since $m$ is the minimum such index, and $m<kn-k+2$,
    \[a_{m-n+1} + \dots + a_{m-1} + a_m = (k-2)+a_{(k-1)n+1}+\dots + a_m = k-1\]
    Then, we have the following chain of inequalities,
    \begin{align*}
        k-1 = a_{m-n+1} + \dots + a_{m-1} + a_m &< a_{m+1}+\dots + a_{kn} + a_{kn+1} + \dots + a_{m+n}\\
        &< a_{m+n+1} + \dots + a_{(k+1)n} + a_{(k+1)n +1} + \dots + a_{m+2n}\\
        &\vdots\\
        &< a_{n(n-k)+m+1} + \dots + a_{n^2} + a_{n^2+1} + \dots + a_{n(n-k+1)+m}
    \end{align*}
    Note that since these are strict inequalities the value of each sum must be atleast 1 more than the previous (this is what we use below).\\

    Now, we require $a_{m+1}+\dots + a_{kn} = k-2$ as $a_{(k-1)n+1} + \dots + a_{m+1}+\dots + a_{kn}=k-1$. This then gives us that
    \[a_{kn+1} + \dots + a_{m+n} > 2\]
    Since the sum $a_{kn+1} + \dots + a_{(k+1)n}=k$ again this gives
    \[a_{m+n+1} + \dots + a_{(k+1)n} < k-2\]
    Again using the sum of $a_{(k+1)n+1}+\dots + a_{(k+2)n} = k+1$ we must have
    \[ a_{(k+1)n +1} + \dots + a_{m+2n} > 3\]
    we then continue likewise. If it is indeed possible to have $a_m=1$, then there must exist possible values for all $a_i$ when $a_m=1$. Thus, this continues until the last bath of terms, giving us that
    \[a_{n^2} + a_{n^2+1} + \dots + a_{n(n-k+1)+m} > n-k+2\]
    But, note that since $m=kn-n+p$, we also have
    \[a_{n^2} + a_{n^2+1} + \dots + a_{n(n-k+1)+m} < 1(m-nk+n)+1 = kn -n + p -nk + n +1 = p+1 \leq n-k+2\]
    which is a clear contradiction to the previous inequality. Thus, clear there cannot exist such $c$.\\

Now, by the nature of the claim it is clear that indeed the above described sequence is the only one which works.


What do you think about Tibet?

[img]http://www.tribuneindia.com/2007/20070629/spr6.jpg[/img]

Propaganda...FYI there's already a discussion thread in Round Table, so try avoid discussing it here.


function
algebra
polynomial
algebra unsolved

The function $f(n)$ satisfies $f(0)=0$, $f(n)=n-f \left( f(n-1) \right)$, $n=1,2,3 \cdots$. Find all polynomials $g(x)$ with real coefficient such that
\[ f(n)= [ g(n) ], \qquad n=0,1,2 \cdots \]
Where $[ g(n) ]$ denote the greatest integer that does not exceed $g(n)$.

Let $ \alpha$ be the positive root of $ x^{2}-x-1=0$. Then $ \alpha=1+\frac{1}{\alpha}$ and $ \frac{1}{\alpha^{2}}+\frac{1}{\alpha}=1$
Firstly we prove that $ f(n)=\left[\frac{n+1}{\alpha}\right]$. We use induction. The first few values can be easily verified.
Suppose that it is true for any $ k\leq n-1$. Then $ f(n-1)=\left[\frac{n}{\alpha}\right]$. Let $ k=\left[\frac{n}{\alpha}\right]$. We must prove that $ \left[\frac{n+1}{\alpha}\right]+\left[\frac{f(n-1)+1}{\alpha}\right]=n$, or $ \left[\frac{n+1}{\alpha}\right]+\left[\frac{k+1}{\alpha}\right]=n$.
[b]First case[/b]. $  k<\frac{n}{\alpha}<\frac{n+1}{\alpha}<k+1$.
Then we must prove that $ E=k+\left[\frac{k+1}{\alpha}\right]=n$. It is clear that $ E$ is an integer. It is sufficient to prove that $ E>n-1$ and $ E<n+1$.
Indeed, we have $ k+\left[\frac{k+1}{\alpha}\right]>k+\frac{k+1}{\alpha}-1=k\alpha+\frac{1}{\alpha}-1$. Then $ k\alpha+\frac{1}{\alpha}-1>n-1$ becomes $ k\alpha+\frac{1}{\alpha}>n$ or $ k+\frac{1}{\alpha^{2}}>\frac{n}{\alpha}$. Adding $ \frac{1}{\alpha}$ to both sides we get $ k+1>\frac{n+1}{\alpha}$, which is true by our assumption. For $ E<n+1$ we use $ E<k+\frac{k+1}{\alpha}=k\alpha+\frac{1}{\alpha}<n+1$ since each term of the LHS is smaller than the respective term of the LHS.
[b]Second case[/b]. $ k<\frac{n}{\alpha}<k+1<\frac{n+1}{\alpha}$.
We have to show that $ E=k+1+\frac{k+1}{\alpha}=n$. As in the first case, $ E$ is an integer and we prove that $ n-1<E<n+1$.
Indeed $ E<k+1+\frac{k+1}{\alpha}=(k+1)\alpha<n+1$, or $ k+1<\frac{n+1}{\alpha}$, which is true from our assumption.
Also $ E>k+1+\frac{k+1}{\alpha}-1=(k+1)\alpha-1>\frac{n}{\alpha}\cdot\alpha-1=n-1$, also from our assumption.

Now, clearly $ g$ has degree 1. Let $ g=ux+v$. Since $ f(n)=\left[\frac{n+1}{\alpha}\right]=\left[g(n)\right]$, we get $ \left|n\left(\frac{1}{\alpha}-u\right)+\frac{1}{\alpha}-v\right|<1$. Clearly $ u=\frac{1}{\alpha}$. Then using the density of $ \left\{\frac{n}{\alpha}\right\}$ in $ (0,1)$ we see that $ v=\frac{1}{\alpha}$.
So $ g(n)=\frac{n+1}{\alpha}$.


Find all functions $ f: R\rightarrow R$ such that
$ f(x\plus{}yf(x))\equal{}f(x)\plus{}xf(y)\quad\forall x,y\in R$.

[quote="kyungissac"]Find all functions $ f: R\rightarrow R$ such that
$ f(x \plus{} yf(x)) \equal{} f(x) \plus{} xf(y)\quad\forall x,y\in R$.[/quote]

Posted 4 days ago in the same forum : see http://www.mathlinks.ro/Forum/viewtopic.php?t=295583


Steve goes on a 40-km bike ride. He covers the first half of the distance averaging a speed of 15 km/hr. In order to average 20 km/hr for the entire trip, how many kilometers per hour must his average speed be during the second half of the trip?

If he covers the first 20 km of the bike ride at a speed of 15 km/hr, it takes him 80 minutes. He wants to average a speed of 20 km/hr. Let $ x$ be the number of minutes it takes him to travel the second half of the trip.

$ \frac{40}{80 \plus{}x} \equal{} \frac{20}{60}$

$ 2400 \equal{} 1600 \plus{} 20x$

$ 40 \equal{} x$

So his speed during the second half of the trip allowed him to travel $ 20$ km in $ 40$ minutes. That's a speed of $ \boxed{30}$ km/hr.


geometry
inequalities
geometric transformation
reflection
triangle inequality

[hide="Solution"]

Consider three points in space.  Let $A=(2,-3,3), B=(-3,1,-1),$ and $C=(x,y,0)$.  We want to find the minimum value of $AC+BC$.  This occurs when $C$ is on $\overline{AB}$.  So the minimum value is just the length of $AB$ which is $\sqrt{57}.$[/hide]

[quote="Altheman"][quote="warut_suk"]Find the minimum value of
$\sqrt{(x-2)^2+(y+3)^2+9}+\sqrt{(x+3)^2+(y-1)^2+1}$
where $x,y$ are real numbers[/quote]

[hide="solution"]let $u=<x-2,y+3,3>$ and $v=<-x-3,-y+1,-1>$

by the triangle inequality,

$|u|+|v|\ge |u+v|$ (with equality when they are scalar multiples of each other)

since equality occurs, $\sqrt{5^2+4^2+2^2}=3\sqrt{5}$[/hide][/quote]

Your $2^2$ should actually be a $(3-(-1))^2 = 4^2$, which matches joml88's answer.

More thinking: why can't we use the points $(2, -3, 3)$ and $(-3, 1, 1)$ to obtain a minimum value of $3\sqrt{5}$?

I should've explained that in my solution.  You can have either $(-3,1,1)$ or $(-3,1,-1).$  What's the difference, eh?  Well, recall that we want the minimum value of $AC+BC$.  This happens to be length of $AB$ iff and $AB$ passes through the $z$-plane.  That's why you want to take $(-3,1,-1)$.  (You could take the other one, but then you would have to find the shortest route from $A$ to the $z$-plane and then to $B$...which you would do by considering the reflection of $B$ across the $z$-plane anyways...)

[quote="Altheman"]$|u|+|v|\ge |u+v|$ (with equality when they are scalar multiples of each other)[/quote]But that scalar must be positive.


Here are a few problems for you to try, some are more challenging than others. Can you get the solutions?

[b]Geometry:[/b]

1.) Rectangle ABCD has adjacent sides of lengths 7 and 24. What is the shortest distance between vertex A, and diagonal BD.

2.) An isoceles right triangle ABC is given with right angle B. Point D is on side BC. Angle bisector AD has length 4. Determine the area of triangle ABC.

[b]Algebra:[/b]

1.) Andrew gave Katie a 5-yard head start in a 100-yard dash and Andrew lost by only $\frac{1}{4}$ of a yard. In how many [u]more[/u] yards, would Andrew have overtaken Katie?

[b]Other:[/b]

1.) How many four digit numbers can be formed using the digits 1, 2, 3, 4, 5, and 6, if the number is odd, between 3000 and 5000, and digits can be repeated?

2.) Find the number of positive integers less than 1000 in which the sum of the digits is equal to 8. (For example, 71, 521, 440)

Good Luck!

[quote="Mario_KaRt"][b]Geometry:[/b][/quote]

[quote="Mario_KaRt"]1.) Rectangle ABCD has adjacent sides of lengths 7 and 24. What is the shortest distance between vertex A, and diagonal BD.[/quote]

[hide="Solution"]BD's length is 25, because 7-24-25 is pythagorean triple. The area of ABD is 7*24/2=7*12=84. The area is also [shortest distance from A to BD]*BD/2. So 168=[length we want to find]*25. So the answer is 168/25.[/hide]

[quote="Mario_KaRt"]2.) An isoceles right triangle ABC is given with right angle B. Point D is on side BC. Angle bisector AD has length 4. Determine the area of triangle ABC.[/quote]

[hide="Solution"]Letting $AB=x$ and $BD=y$, we have $AC=x\sqrt{2}$ and $\frac xy = \frac{x\sqrt{2}}{x-y}$, so $y=\frac{x-y}{\sqrt{2}}$; $y\sqrt{2}=x-y$;
$y+y\sqrt{2}=x$; $y=\frac{x}{\sqrt{2}+1}=x(\sqrt{2}-1)$. So $AD=\sqrt{x^2+y^2}=\sqrt{x^2+x^2(3-2\sqrt{2})}=\sqrt{4x^2-2x^2\sqrt{2}}=4$. So $4x^2-2x^2\sqrt{2}=16$; $x^2(4-2\sqrt{2})=16$; $x^2=\frac{16}{4-2\sqrt{2}}=\frac{8}{2-\sqrt{2}}$; $\frac{x^2}{2}=\frac{4}{2-\sqrt{2}}=\frac{8+4\sqrt{2}}{2}=4+2\sqrt{2}$,the answer.[/hide]

[quote="Mario_KaRt"][b]Algebra:[/b][/quote]

[quote="Mario_KaRt"]1.) Andrew gave Katie a 5-yard head start in a 100-yard dash and Andrew lost by only $\frac{1}{4}$ of a yard. In how many [u]more[/u] yards, would Andrew have overtaken Katie?[/quote]

[hide="Solution"]Andrew ran 99.75 yards in the same time Katie ran 95 yards. Let Andrew's speed be x and Katie's speed be y. So 99.75/x=95/y and x/y=99.75/95. We want to find z such that z/x=(z-.25)/y. So z/(z-.25)=4z/(4z-1)=x/y=99.75/95=21/20. So 4z=21 and z=5.25.[/hide]

[quote="Mario_KaRt"][b]Other:[/b][/quote]

[quote="Mario_KaRt"]1.) How many four digit numbers can be formed using the digits 1, 2, 3, 4, 5, and 6, if the number is odd, between 3000 and 5000, and digits can be repeated?[/quote]

[hide="Solution"]To be odd the number must end in 1,3,or 5. To be between 3000 and 5000, the first digit must be 3 or 4. We have six choices for the other digits. 2*6*6*3=216.[/hide]

[quote="Mario_KaRt"]2.) Find the number of positive integers less than 1000 in which the sum of the digits is equal to 8. (For example, 71, 521, 440)[/quote]

[hide="Solution"]For one-digit numbers, there is 1.
For two-digit numbers, there are 8.
For three-digit numbers, there are 36.
1+8+36=45.[/hide]

[quote="Mario_KaRt"]Good Luck![/quote]


limit
real analysis
real analysis unsolved

Hi,

I think this limit is nice:

$\lim_{n\rightarrow \infty}\left(\frac{\sum_{j=1}^{n+2}j^{j}}{\sum_{j=1}^{n+1}j^{j}}-\frac{\sum_{j=1}^{n+1}j^{j}}{\sum_{j=1}^{n}j^{j}}\right)$

I got the value of this limit as $e$, assuming that I made no computational mistakes,  however my method is long and computational so I was wondering whether there is a technique or a trick that can be used for calculating this limit. I am looking for a non standard method...
Didi


For $ a;b;c >0$ and $ a\plus{}b\plus{}c\equal{}1$
Prove that $ \frac{{\sqrt {a^2  \plus{} abc} }}{{c \plus{} ab}} \plus{} \frac{{\sqrt {b^2  \plus{} abc} }}{{a \plus{} bc}} \plus{} \frac{{\sqrt {c^2  \plus{} abc} }}{{b \plus{} ac}} \le \frac{1}{{2\sqrt {abc} }}$

[quote="Mr.Dung"]For $ a;b;c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$
Prove that $ \frac {{\sqrt {a^2 \plus{} abc} }}{{c \plus{} ab}} \plus{} \frac {{\sqrt {b^2 \plus{} abc} }}{{a \plus{} bc}} \plus{} \frac {{\sqrt {c^2 \plus{} abc} }}{{b \plus{} ac}} \le \frac {1}{{2\sqrt {abc} }}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204969


I've got this problem to do and I can't figure it out at all.

(csc²x)(secx) = (secx) + (cscx)(cotx)

I've got to verify that the left side equals the right side, without doing anything to the right side of the equation.

I'm sorry that it's not in LaTeX.  I've downloaded it but I haven't completely figured out how to use it.

$(\csc^{2}x)(\sec x)=\frac{1}{\sin^{2}x\cos x}$
$\Longrightarrow \frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos x}=\frac{1}{\cos x}+\frac{\cos x}{\sin^{2}x}$
$\Longrightarrow \sec x+(\csc x)(\cot x)$

$\mathbb{QED}$

You don't need to download $\LaTeX$ to use it on the forums.  It's already there.  You just need to put dollar signs around your math, which will cover rather a lot.  The rest you can learn either from the site's $\LaTeX$ section, or by hovering over other people's $\LaTeX$.

This link can also help you:
[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]


floor function
function
algebra
polynomial

Determine the number of $ 8$-tuples of nonnegative integers $ (a_1,a_2,a_3,a_4,b_1,b_2,b_3,b_4)$ satisfying $ 0\le a_k\le k$, for each $ k \equal{} 1,2,3,4$, and $ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} 2b_1 \plus{} 3b_2 \plus{} 4b_3 \plus{} 5b_4 \equal{} 19$.

[hide="Solution"]
Let $ c_k\equal{}a_k \plus{} (k\plus{}1)b_k$

Note that there is only one way to write $ c_k$ in the given form: we must have $ b_k \equal{} \left \lfloor \frac{c_k}{k\plus{}1} \right \rfloor$, and $ a_k \equal{} c_k \minus{} (k\plus{}1) \left \lfloor \frac{c_k}{k\plus{}1} \right \rfloor$.

[i]In other words, $ b_k$ is the number of complete times $ k\plus{}1$ goes into $ c_k$ and $ a_k$ is the remainder of $ c_k$ upon division by $ k\plus{}1$.[/i]

Thus, the number of solutions to the given equation equals the number of nonnegative solutions to:

$ c_1 \plus{} c_2 \plus{} c_3 \plus{} c_4 \equal{} 19$.

By balls and urns, this is $ \binom{19\plus{}4\minus{}1}{4\minus{}1} \equal{} \binom{22}{3} \equal{} \boxed{1540}$
[/hide]

Hint

[hide]Think of Generating Functions, it makes it a heck of al ot easier.[/hide]

Wow, I couldn't get this during the test, but the solution quickly became really obvious a few hours later when I looked at the problem again.  :(

Nice solution Zuton Force.

[hide="Combining the above two approaches"] The relevant generating function is

$ (1 \plus{} x)(1 \plus{} x \plus{} x^2)(1 \plus{} x \plus{} x^2 \plus{} x^3)(1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4)(1 \plus{} x^2 \plus{} x^4 \plus{} ...)(1 \plus{} x^3 \plus{} x^6 \plus{} ...)(1 \plus{} x^4 \plus{} x^8 \plus{} ...)(1 \plus{} x^5 \plus{} x^{10} \plus{} ...)$

which is clearly equal to $ (1 \plus{} x \plus{} x^2 \plus{} ...)^4$.  :) [/hide]

Wait, but if we're given bounds on the variables, how can we use the infinite generating functions (since these bounds restrict the possibilities).

We're given bounds on $ a_k$ which correspond to the polynomial factors in the beginning.  There are no such restrictions on $ b_k$ :)


calculus
inequalities proposed
inequalities

Let $ a,b,c>1: a\plus{}b\plus{}c\equal{}abc$. Find min of: $ \frac{b\minus{}2}{a^2}\plus{}\frac{c\minus{}2}{b^2}\plus{}\frac{a\minus{}2}{c^2}$

hello, with the help of calculus we get the minimum as $ \sqrt{3}\minus{}2$ which will be attained for $ a\equal{}b\equal{}c\equal{}\sqrt{3}$.

[quote="Dr Sonnhard Graubner"]hello, with the help of calculus we get the minimum as $ \sqrt {3} \minus{} 2$ which will be attained for $ a \equal{} b \equal{} c \equal{} \sqrt {3}$.[/quote]

OK, it is $ \sqrt3\minus{}2$, but what about solution?


inequalities
algebra
polynomial
inequalities unsolved

try to solve this important inequality. it's often use to find max and min when we cant use nolder cauchy and other inequalitys.

I don't think we can really find "nice" conditions on the coefficients... but the following theorem may be related-
A polynomial $ P(x)$ with real coefficients is non-negative iff it's a sum of two squares of polynomials.

Here's a proof (not mine):
Let $ P(x)\equal{}\sum_{i\equal{}0}^{n} a_{i}x^{i}$. If $ P(x)\equal{}G(x)^{2}\plus{}H(x)^{2}$, then P is non-negative.
If P is non negative, it must be of even degree. If $ a\plus{}bi$ is a root of P, then so is $ a\minus{}bi$. Also, each real root is of even multiplicity: let $ x_{1} \in R, P(x_{1})\equal{}0$. If the root is of odd multiplicity $ 2k\minus{}1$, then: $ P(x)\equal{}(x\minus{}x_{1})^{2k\minus{}1}F(x)$. WLOG we can assume that $ F(x_{1})>0$, and then $ P(x)$ is negative when x tends to $ x_{1}$ from below, a contradiction.
Therefore, we can represent P in the following way: $ P(x)\equal{}a_{n}\prod_{i\equal{}1}^{\frac{n}{2}} (x\minus{}x_{i})(x\minus{}\overline{x_{i}})$
$ T(x)\equal{}\prod_{i\equal{}1}^{\frac{n}{2}} (x\minus{}x_{i})\equal{}R(x)\plus{}iS(x)$, where R and S are polynomial with real coefficients.
hence, $ P(x)\equal{}a_{n}(R(x)\plus{}iS(x)\overline{(R(x)\plus{}iS(x)}\equal{}(\sqrt{a_{n}}R(x))^{2}\plus{}(\sqrt{a_{n}}S(x))^{2}$ ($ a_{n}$ is obviously positive).
I'm not sure it helps much. But I also don't think there's a satisfying answer... We need to check the multiplicity of all the roots, and maybe compare coefficients of P and $ G^{2}\plus{}H^{2}$, but there are usually too many options... Anyway, the above theorem is really nice  :blush:


What is the complexity of deciding whether a given number of axis-parallel unit squares can be packed into a simple polygon (without holes)?

order by $(x_i, y_i)$ lexicographically and draw all the boxes in that order,
this way each time you draw a box it connects at most 2 other previous boxes. If they are in diferent coneted components, then it doesn't form a cicle (a void space). Otherwise, if they are from the same component, either they are in a block (wich is easy to verify) or they form a cicle (then it form a whole hole)


Very nice.I am so happy that I can figure out the solution ;)
So here it is.
Solution:
with $(a+b+c+d)^2=0$
then:$a^2+b^2+c^2+d^2+2(ab+bc+cd+da+bd+ac)=0$
thus:
$(ab+bc+cd+da+bd+ac)^2 + 12 \geq 6(abc+abd+bcd+dca)$
$\iff (a^2+b^2+c^2+d^2)^2 + 48 \geq 24(abc+abd+bcd+dca)$
$\iff (a^2+b^2+c^2+(a+b+c)^2)^2 + 48 \geq 24(abc-(a+b+c)(ab+bc+ca))$(use $d=-(a+b+c)$)
$\iff \left((a+b)^2+(b+c)^2+(c+a)^2 \right)^2+48 \geq -24(a+b)(b+c)(c+a)$
$\iff \left(x^2+y^2+z^2 \right)^2+48 \geq -24xyz$(let $x=b+c$ and $y,z$ similar)
following I will prove:
$\left(x^2+y^2+z^2\right)^2+48 \geq 24|xyz|$  $(1)$
let $t=\sqrt[3]{|xyz|}$
then we have:
$\left(x^2+y^2+z^2\right)^2+48 \ge 9t^4+48$
so to prove inequality $(1)$ we should only prove:
$3t^4+48 \ge 8t^3$
but we have:$3t^4+48 -8t^3=(t-2)^2(3t^2+2t+2) \ge 0$
inequality $(1)$ is prove.
Obviously the original inequality is still proved.
PS:when does the equality hold :?: find it yourself ;)

Actually Cefer(first of first together with other romanian guy)  from  our(azerbaijan) team solved like this.

My solution was similar to yours :)
  $(x^2+y^2+z^2)^2+48 \geq 3( x^2y^2+y^2z^2+z^2x^2+16) \geq 3*4*2*|xyz| \geq -24xyz$
  Two guys from Romania and Azerbaijan took 31 points.

I have another solution.

Just use Following ineqs.

$a^2b^2 + b^2c^2 +c^2a^2 +1 \geq 4abc$

$2abc(a+b+c) +2 \geq 8abc$

These are just using $AM-GM$.

[quote="Chang Woo-JIn"][quote="zhaobin"]so to prove inequality $(1)$ we should only prove:
$3t^4+48 \ge 8t^3$[/quote]
You're wrong with this part. It should be $3t^4 +16\geq 8t^3$.

And $\Leftrightarrow (t-2)^2 (3t^2+4t+4)\geq0$.[/quote]
you are right,thanks,
but I can't understand your proof :(
edit:I know your proof,it is just for the rest part ;)

That is one of the six questions in Viet Nam competition ín1999

yes really nice and  who is from Turkish high school ?

[quote="Mamat"]I study in Kyrgyz-Turkish High School[/quote]

Great me too but  i in the Tajik - Turkish high school will did (you participate) in the IMO ?

I'm also studying at Baku(Azerbaijan) Private Turkish High School.(last grade) ;)

[quote="Mamat"]If $a+b+c+d=0$ then prove that
 $(ab+bc+cd+da+bd+ac)^2 + 12 \geq 6(abc+abd+bcd+dca)$
 By the way only a few people could solve it[/quote]

I've got a nice proof (using analysis):
Let
$\displaystyle a \geq b \geq c \geq d \Rightarrow \\ P(x)=x^4+(a+b+c+d)*x^3+(ab+ac+{.\;.\;.\;}+cd)*x^2+(abc+{.\;.\;.\;})x+abcd=(x+a)(x+b)(x+c)(x+d) \\ \Rightarrow Q(x)=P(x){'\;}=4(x+k)(x+l)(x+m) {\; for\; some\; x,\; y,\; z,\; such\; as\;} \\ a \geq k \geq b \geq l \geq c \geq m \geq d \\ \Rightarrow \\ 4(k+l+m)=3(a+b+c+d)=0 \\ 4(kl+km+lm)=2(ab+{.\;.\;.\;}) \\ 4klm=(abc+abd+acd+bcd)$
So we can assume l= -m -k, where m<0 , k<0 (In the enother case the inequality is clear :   ...>=0>=...) ;  So we have to prove
$\displaystyle (m^2+k^2+km)^2+3 \geq 6km(k+m)$
And this inequality isn't hard  - we can use the Caushi's inequality several times (if k=m then the parts are equal)
 :)

A plane of another solution:
The fact, which is proved with help analysis:
If equation $x^4+kx^2+lx+m=0$ has 4 roots($-a,-b,-c,-d$), then $8k^3\geq 27l^2$.
But then $k^2+12=\frac{k^2}{3}+\frac{k^2}{3}+\frac{k^2}{3}+12\geq 4\sqrt{\frac{2k^3}{3}}\geq 6l$(using the fact).

[quote="spider_boy"]I'm also studying at Baku(Azerbaijan) Private Turkish High School.(last grade)[/quote]

 Hello!!!  Did you participate in Jautykov Olimpiad?

[quote="Davron"]..........school will did (you participate) in the IMO ?  [/quote]
 I believe that I will :)

[quote="Mamat"][quote="spider_boy"]I'm also studying at Baku(Azerbaijan) Private Turkish High School.(last grade)[/quote]

 Hello!!!  Did you participate in Jautykov Olimpiad?    Yes!

[quote="Davron"]..........school will did (you participate) in the IMO ?  [/quote]
 I believe that I will :)[/quote]

[quote="Chang Woo-JIn"]I have another solution.

Just use Following ineqs.

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} 1 \geq 4abc$

$ 2abc(a \plus{} b \plus{} c) \plus{} 2 \geq 8abc$

These are just using $ AM \minus{} GM$.[/quote]AM-GM is only valid for positive integers.

[quote="Johan Gunardi"]AM-GM is only valid for positive integers.[/quote]
For non-negative reals too.  :wink:


let $ d\equal{}a^{1999}\plus{}b^{1999}\plus{}c^{1999}$,  where a,b,c  are integers  such that $ a\plus{}b\plus{}c\equal{}0$

a) is it possible that d=2

b) is it possible that d is prime ?

d is not prime,by fermat's theorem 1999 divides the expression.
(1999 is a prime)


algebra
polynomial
LaTeX
induction
algebra unsolved

Let be given a polynomial $P(x)=x^2-1$.How many real numbers $x$ satisfying this equation:
$P(P(...(P(x))...))=0$($2006$ $P$'s)

I think it isn't easy like that you say,we can imagine that for $P(P(\ldots P(x)\ldots))=0(nP\mbox{'s})$ there are $a_n$ real root and we can get a sequence and solve it.

In n-th step (n>2) we have all roots (n-2) th step and 2 roots +-sqrt(1+sqrt(1+...)) with n  1. Therefore a(n)=a(n-2)+2, a(1)=2,
a(2)=3 {-sqrt(2),0,sqrt(2)}. Therefore a(2006)=2007.

[quote="Rust"]In first solution Sqrt(2006),sqrt(20004)  means sqrt - 2006 times with 2006 1, sqrt -2004 times with 2004 1, sqrt(1+sqrt(1+sqrt(1+sqrt(1+....[/quote

For god's sake, Rust use $Latex$.

For some reason, Rust stubbornly refuses to use LaTeX - I guess netiquette means nothing to him.

[quote="Rust"]In n-th step (n>2) we have all roots (n-2) th step and 2 roots +-sqrt(1+sqrt(1+...)) with n  1..[/quote]
Are you sure these are [u][b]all[/b][/u] roots?

Yes.
They are may be expressed by recurent formuls. Let P(0,x)=x, P(1,x)=x^2-1,P(n+1,x)=P(1,P(n,x)), x(0)=0,x(n+1)=sqrt(1+x(n)).
We have P(n,x(n))=0, if P(n,x)=0, then P(n-1,P(1,x))=0. Therefore real roots P(n,x) can be expessed +-sqrt(1+y),were y is roots P(n-1,x), or 0 if (P(n-1,-1)=0. By induction proved all roots P(n,x) are +-x(n),+-x(n-2),+-x(n-4),...

[b]Rust[/b] is right [b]James[/b],  but the number of solutions is 4012
Actually, all the roots of $P_{2006}(x)$ are $a_1\cdot\sqrt{1+a_2\cdot\sqrt{1+\cdots +a_{2006}}}$ where $a_i=1$ or $a_i=-1$
But if $a_k=-1$ then $a_{k+1}=a_{k+2}=\cdots =a_{2006}=-1$
So there are 2006 cases for $(a_2,a_3,...,a_{2006})$ and 2 cases for $a_1$ ( to be 1 or -1 )
Hence the number of solutions is 2006x2=4012


function
real analysis
real analysis unsolved

Can you use conditionally convergent series in your computations and actually derive results from them? (like the alternating harmonic series)

Please be more precise? What do you mean? Something other than ascertaining convergence?

Consider the alternating harmonic series which is In2. The Riemann series theorem states that by rearranging the terms of a conditionally convergent serie you can make it converge to any desired amount or even diverge. So if you have a bound on a function (for instance) which consists of this conditionally convergent serie then can you put In2 instead of it such that your bound is still right?

Of course; conditionally convergent series still converge.  You just can't do things like add two series by adding their terms pairwise.


modular arithmetic
combinatorics unsolved
combinatorics

let A={1,2,...,n}  and a(1),a(2),...,a(n) be subsets of A wich

1) la(i)l <n
2)for every  0<i,j<n+1      a(i)  :cap: a(j)=t   wich t is constant.

prove that  la(1)l=la(2)l=......=la(n)l.

Are you sure?
Consider $A_i=\{i,n\}$, $i=1,...,n-1$, $A_n=\{1,2,...,n-1\}$.

For similar problems look at http://www.mathlinks.ro/Forum/viewtopic.php?t=32426 and http://www.mathlinks.ro/Forum/viewtopic.php?t=32408

sorry!i forget the third condition.
3)a(i) :cap: {i} is null

I can define $A_i=\{i\pmod{n-1}+1,n\}$, so $A_i\cap \{i\}=\varnothing$.

you are right
i saw this problem in a magazing.but...


logarithms
number theory unsolved
number theory

Prove that there exist infinitely many positive integers $n$ such that $n!$ is a multiple of $n^{2}+1$

There is still an unsolved problem on mathlinks: if $a>0$ there are infinitely many $n$ for which $n^{2}+1$ divides $[an]!$.  :wink:

Isn't the solution exactly the same? Choose $k_{1}$ such that $2k_{1}^{2}+2k_{1}+1$ is divisible by some very big prime $p_{1}$. Then choose $k_{2}$ such that $2k_{2}^{2}-2k_{2}+1$ is divisible by even larger prime $p_{2}$. Then choose a really huge $k$ such that $k\equiv k_{j}\mod p_{j}$ for $j=1,2$ and finish the problem the same way as before :?

So, let's put $a$ to a different position. :) Prove or disprove: for every $a>0$ there exist infinitely many $n$ such that $n^{2}+1$ divides $[n^{a}]!$. The above trick allows us to prove it for $a>1/2$ or something like that.

This assertion is also true, but I need to use a result of Schinzel (Acta Arithm., Warszawa, t. 13, 1967, p. 177-236): there are infinitely many $n$ for which $P(n^{2}+1)<C\frac{\ln n}{\ln(\ln (\ln n))}$ for some constant $C$. It is not difficult to see that for a big such number $n^{2}+1$ divides $[n^{a}]!$. Here $P(x)$ is the largest prime factor of $x$. I'd like however to see a more elementary approach...


group theory
superior algebra
superior algebra unsolved

If  $ c \equal{} (1$ $ 2 ... n)$ is an n-cycle and $ t \equal{} (1$  $ d)$ is a transposition, for which $ d$
do $ t$ and $ c$ generate the whole symmetric group of order $ n$? ($ Sym \equal{} <c , t>$)

any transposition and any $ n$-cycle can generate $ S_n$.

(123456) and (13) don't generate the symmetric group of 6 elements

if and only if $ gcd(d\minus{}1,n)\equal{}1$.


algebra
polynomial
number theory unsolved
number theory

Find all positive integers $n$ for which all coefficients of polynomial $P(x)$ are divisible by $7,$ where
\[P(x) = (x^2 + x + 1)^n - (x^2 + 1)^n - (x + 1)^n - (x^2 + x)^n + x^{2n} + x^n + 1.\]

Sorry, for example $2|a^{2}+a$ for any integer $a$, but the coefficients are not divisible by $2$...

TomciO is right it says its coefficients are divisible by 7

Oh, okay, I made a mistake...
But at least the answer must be $1 (mod 6)$ or $5 (mod 6)$..
I'll try to complete the solution, thx for correcting..

It is easy to show, that $P_{n}(x)=\sum_{m}a_{m}x^{m}$ had symmetric koefficients $a_{k}=a_{2n-k}$. Therefore sufficiently calculate $a_{m}, m\le n$. We have \[a_{m}=\sum_{k=[m/2]+1}^{m-1}C_{n}^{k}C_{k}^{m-k}.\]
It give $a_{0}=a_{1}=a_{2}=0, a_{3}=n(n-1),a_{4}=\frac{n(n-1)^{2}}{2},...$
It give $7|P_{n}(x)$ if and only if $7|n(n-1)$.

i dont think it was the only necessary relation
just look at the second post

answer is $7^{k}$ and $7^{k}+7^{l}$ such that $k$ and $l$ are integers, $k,l>-1$

We claim that the only possibilities are integers of the form $7^k$ or $7^k + 7^\ell$, where $k, \ell \in \mathbb{Z}_{\ge 0}.$

Let us first show that no other numbers work.

Firstly, define $P_n(x)$ to be $(x^2+x+1)^n - (x^2+1)^n - (x+1)^n - (x^2+x)^n +x^{2n}+x^n+1$ and $Q_n(x) = P_n'(x)$ for each $n \in \mathbb{N}.$ Also define $R_n(x) = (x^2+x+1)^{n} - (x^2+1)^{n} - (x^2+x)^{n} + (x^2)^{n}$ for each $n \in \mathbb{N}.$ Then, we have that:

$$\frac{Q_n(x)}{n} = (x^2+x+1)^{n-1}(2x+1) - (x^2+1)^{n-1}(2x) - (x+1)^{n-1} - (x^2+x)^{n-1}(2x+1) + 2x^{2n-1} + x^{n-1}. \qquad (1)$$

Call a positive integer $n$ [b] tasty [/b] if $P_n(x)$ has all coefficients divisible by $7$.

[b] Lemma 1. [/b] If $n$ is tasty, then either $n = 2$ or $n \equiv 0, 1$ (mod $7$).

[hide = Proof]

If $n \le 3$ then the lemma is trivially true. Else, if $n \ge 4$, then we analyze the $x^3-$coefficient in $P_n(x).$ It is actually just $n(n-1)$, so $7|n(n-1)$ proves the lemma.

[/hide]

[b] Lemma 2. [/b] If $n$ is a multiple of $7$, then $n$ is tasty if and only if $\frac{n}{7}$ is.

[hide = Proof]

This is obvious from the Frobenius Endomorphism.

[/hide]

The following two lemmas are proven in analogous ways as Lemmas $1, 2$, and so the proofs are left as exercises to the reader.

[b] Lemma 3. [/b] If $R_n(x)$ has all coefficients divisible by $7$, then $n \equiv 0, 1$ (mod $7$).

[b] Lemma 4. [/b] If $n$ is a multiple of $7$, then $R_n(x)$ has all coefficients divisible by $7$ if and only if $R_{\frac{n}{7}}(x)$ does.

The following lemma will later prove very useful.

[b] Lemma 5. [/b] If $R_n(x)$ has all coefficients divisible by $7$, then $n$ is a power of $7$.

[hide = Proof]

From Lemmas $3$ and $4$, it suffices to show that if $n \equiv 1$ (mod $7$) is so that $R_n(x)$ has all coefficients divisible by $7$, then $n = 1.$ Let's show this.

Notice that $R_n'(x) = (x^2+x+1)^{n-1}(2x+1) - (x^2+1)^{n-1} (2x) - (x^2+x)^{n-1} (2x+1) + (x^2)^{n-1} (2x).$ Hence, because $7|n-1$, by Frobenius Endomorphism and looking at only the terms which have exponents $0$ (mod $7$), we have that

$$(x^2+x+1)^{n-1} - (x^2+x)^{n-1}$$

has all coefficients divisible by $7$.

If $n - 1 > 0$, then plugging in $x = 0$ gives an immediate contradiction. Hence $n = 1$ and the lemma is proven.

[/hide]

Call a positive integer $n$ [b] good [/b] if $Q_n(x)$ has all coefficients divisible by $7$. The following is the key claim.

[b] Claim. [/b] A positive integer $n \equiv 1$ (mod $7$) is good if and only if $n-1$ is a power of $7$.

[hide = Proof]

From $(1)$, we have modulo $7$ that:

$$(x^2+x+1)^{n-1}(2x+1) - (x^2+1)^{n-1}(2x) - (x+1)^{n-1} - (x^2+x)^{n-1}(2x+1) + 2x^{2n-1} + x^{n-1} \equiv 0.$$

From the Frobenius Endomorphism, we know modulo $7$ that:

$$(x^2+x+1)^{n-1}(2x) - (x^2+1)^{n-1}(2x) - (x^2+x)^{n-1}(2x) + 2x^{2n-1} \qquad (2)$$

has all nonzero terms with exponents $1$ (mod $7$) and

$$(x^2+x+1)^{n-1}(1 - (x+1)^{n-1} - (x^2+x)^{n-1}(1) + x^{n-1}$$

has all nonzero terms with exponents $0$ (mod $7$). Therefore, it's clear that both must be identically $0$ modulo $7$, and so dividing $(2)$ by $2x$ gives that, modulo $7$, we have:

$$(x^2+x+1)^{n-1} - (x^2+1)^{n-1} -(x^2+x)^{n-1} + (x^2)^{n-1} \equiv 0,$$

i.e.,

$$R_{n-1}(x) \equiv 0.$$

From Lemma $5$, this implies that $n-1$ is a power of $7$ and so the lemma is easily proven.

[/hide]

[b] Lemma 7. [/b] If $n \equiv 1$ (mod $7$) is tasty, then it is either $1$ or of the form $7^k+1$ for some $k \in \mathbb{N}.$

[hide = Proof] Suppose that $n \ge 15$ was tasty. We claim that $n = 7^k + 1$ for some $k \in \mathbb{N}_{\ge 2}$, which proves the lemma. Indeed, if $P_n(x)$ has all coefficients divisible by $7$, then so does $P_n'(x) = Q_n(x).$ Hence, $n$ is good, and so by the Claim the lemma is proven.

[/hide]

Hence, from Lemmas $1$ and $7$, we know that all tasty $n$ are either $1, 2,$ multiples of $7$, or of the form $7^k+1$ for $k \in \mathbb{N}.$

By Lemma $2$, we know that if some positive integer $n = 7^x \cdot t$ where $7 \nmid t$ and $x > 0$ is tasty, then $t$ is tasty. Hence, we have either $t = 1$, $t = 2$, or $t = 7^k + 1$ for some $k \in \mathbb{N}.$ This clearly implies that all multiples of $7$ which are tasty are either powers of $7$ or sums of two powers of $7$.

The previous discussion easily shows that all tasty numbers must be of the form $7^k$ or $7^k + 7^\ell$, where $k, \ell \in \mathbb{Z}_{\ge 0}.$

Let's now show that all of these numbers are actually tasty. By Lemma $2$, it suffices only to show that $1, 2,$ and numbers of the form $7^k+1$ for $k \in \mathbb{N}$ are tasty.

It's easy to check by hand that $1, 2$ are tasty, so let's just show that $7^k+1$ is tasty for all $k \in \mathbb{N}.$

Observe that by Legendre's Formula, the only triples $(a, b, c) \in (\mathbb{Z}_{\ge 0})^3$ with $7 \nmid \frac{n!}{a! b! c!}$ are permutations of $(7^k + 1, 0, 0)$ or of $(7^k, 1, 0).$ With this observation, it's quite easy to show that $7^k+1$ is tasty for all $k \in \mathbb{N}.$

We have therefore shown that all positive integers $n$ of the form $7^k$ or $7^k + 7^\ell$, where $k, \ell \in \mathbb{Z}_{\ge 0}$, are indeed tasty, and so hence that is our answer.

$\square$


geometry conics ellipse trigonometry calculus integration parameterization

Calculate the area of the part of the plane $Ax+By+Cz+D=0$ which lies inside the cylinder $x^2+y^2=1$. We assume $C\ne 0$.

The plane and the cylinder determine an ellipse, and its parametric equation is  :
$x=\cos t , y=\sin t , z = a\cos t +b\sin t +d$, where $a=-\frac AC,b=-\frac BC,d=-\frac DC$
The center $K$ of the ellipse is the intersection of the plane and the cylinder's axe $Oz$, so : $K=(0,0,d).$
For all ellipse's point $P$, we have $KP^2 = 1+(a\cos t + b\sin t)^2= 1 + (a^2+b^2)\cos ^2(t-m)$ ,
where $\cos m=\frac a{\sqrt{a^2+b^2}}, \sin m=\frac b{\sqrt{a^2+b^2}}$
Now we can see that the semi-axes of the ellipse are $KP_{min}=1 , KP_{max}=\sqrt{1+a^2+b^2}$

Finally the asked surface is $S = \pi KP_{min}KP_{max}=\pi \sqrt{1+a^2+b^2}= \pi\frac{\sqrt{A^2+B^2+C^2}}{|C|}$
 I hope no mistake ! :cool:

I think an easier approach is too see that the area of the ellipse formed by the intersection of the plane and the cyllinder is the area of the unit circle times the cosine of the angle between the given plane and the $xy$ plane.

An even easier way would be to set up a surface integral over the cylinder.  The $D$ doesn't really matter as it only shifts the plane.  You have $z=-\left(\frac{A}{C}x+\frac{B}{C}y\right)$ over $x^{2}+y^{2}=1$.

$\iint_{S}\,dS=\int_{0}^{2\pi}\int_{0}^{1}\sqrt{1+\left(\frac{A}{C}\right)^{2}+\left(\frac{B}{C}\right)^{2}}\,r\,dr\,d\theta$

$=\pi\sqrt{\frac{A^{2}+B^{2}+C^{2}}{C^{2}}}=\pi\frac{\sqrt{A^{2}+B^{2}+C^{2}}}{\left|C\right|}$


algebra
polynomial
linear algebra
matrix
calculus
derivative
number theory

Find a polynomial of degree 5 p(x) such that p(x)+1 is divisible by (x-1)^3 and p(x)-1 is divisible by (x+1)^3.

p(x)=(ax^2+bx+c)(x-1)^3-1 and p(x)=(dx^2+ex+f)(x+1)^3+1 . Okay, we can see a=d by equating fifth degree coefficients in [p(x)+1]-1 and [p(x)-1]+1. We also see (equating constants) that -c-1=f+1 => f=-c-2. By equating fourth degree coefficients, we have -3a+b=3d+e => 6a=b-e. Equating first degree coeffs gives us -b+3c=e+3f equating third degree coefs gives us now -3b+c=3e+f. these two give us 8b=-8e => b+e=0 and c=f=-1. equating second degree coeffs now gives us -a+3b-3c=a-3b+3c => 6b=2a+6c => 6b=2(-1/9)+6(-1)=-2/9-6=-56/9 => b=-28/27. So we have (a,b,c,d,e,f)=(-1/9,-28/27,-1,-1/9,28/27,-1). and then you do some algebra. ok so i probably made a mistake somewhere. i seem fond of doing that. and there's probably a better way. but i cannae find it.

Your method is right, but I got a different answer. I got [hide](a, b, c, d, e, f) = (-3/8, -9/8, -1, -3/8, 9/8, -1).[/hide]

I believe your way is more or less the shortest. You could possibly solve the system of equations a bit faster using an augmented matrix and row elimination - or by just plugging it into a TI. The only other way I see to solve this is to use the Chinese Remainder Theorem, but that would be a mess.

I figured that if we made p an odd polynomial, then the two conditions become the same condition.  So let p(x) = ax5 + cx3 + e.  I then used the fact that divisibility by (x - 1)3 means that the first three derivatives at 1 are 0.  That led to three linear equations for a, c, and e, namely a + c + e = -1, 5a + 3c + e = 0, and 20a + 6c = 0.  Solving, I got: [hide]a = -3/8, c = 5/4, and e = -15/8[/hide]  So that means that p(x) = [hide]-3/8 x5 + 5/4 x3 -15/8 x[/hide]  I hope I didn't make an error.

I guess calculus is out of scope for an intermediate problem though...

Very nice solution Ravi. My coeffecients agree with yours when you multiply them out. Except I think you meant p(x) = ax5 + cx3 + ex.

Yes, thanks, make that ex, not e.  The rest is still okay.

How can you assume that there is an odd polynomial that satisfies the conditions (and not consider the possibility that there has to be an x^2 and/or x^4 term)?

Well, first, your question said to find a polynomial, and I found one that met at least as strong a condition.

But here's why the polynomial has to be odd.  Note that if polynomial p obeys your conditions, then polynomial -p(-x) also does.  Hence the difference p(x) + p(-x) must be divisible by (x - 1)3 and (x + 1)3.  That means that p(x) + p(-x) must be divisible by the product (x - 1)3 (x + 1)3.  And the only 5th degree polynomial divisible by that product is 0.  Hence p(x) + p(-x) = 0, which means that p is odd.

[quote]Note that if polynomial p obeys your conditions, then polynomial -p(-x) also does. Hence the difference p(x) + p(-x) must be divisible by (x - 1)3 and (x + 1)3. That means that p(x) + p(-x) must be divisible by the product (x - 1)3 (x + 1)3. And the only 5th degree polynomial divisible by that product is 0. Hence p(x) + p(-x) = 0, which means that p is odd.[/quote]

Ok, so you took the difference p(x) + p(-x), which is divisible by (x-1)^3 and (x+1)^3. But what if you instead took the sum p(x) - p(-x), which would also be divisible by (x-1)^3 and (x+1)^3 ? Then, following your reasoning, you would deduce that p(x) - p(-x) = 0 <=> p(x) = p(-x), which would imply that p(x) is even, and not odd. But this is a contradiction, right?

um...i don't think that works. when you subtract, you cancel out those constants (plus or minus 1) because p(x) is not a multiple of those thingies, it is one more or one less than multiples of those thingies. However, when you add, I don't think the constants cancel out, so it's probably not divisible by either (x-1)^3 or (x+1)^3

Oh right, my mistake, I forgot about the +/- 1. Thanks for the help guys.

Here's another solution to the original problem, inspired by gauss's suggestion to use the Chinese Remainer Theorem.  For this particular problem, it's probably overkill, but for higher-degree variants, I think it could be more efficient.

The polynomials (x - 1)3 and (x + 1)3 are relatively prime, which means there are two polynomials a and b such that a(x) (x - 1)3 + b(x) (x + 1)3 = 1.  In particular, we can find a and b using the Euclidean gcd algorithm.  After about 3 steps, I found a(x) = -3/16 x2 - 9/16 x - 1/2 and b(x) = 3/16 x2 - 9/16 x + 1/2.

Now consider the polynomial p(x) = a(x) (x - 1)3 - b(x) (x + 1)3.  (I chose the +1, -1 coefficients because of the constants in the original problem.)  It is easy to see that this p will obey the original conditions, so we're done.


function
calculus
derivative
logarithms
vector
complex analysis
complex analysis unsolved

Suppose that an analytic in some neighborhood of the origin function $f$ admits an analytic continuation along some curve that returns to the origin and the analytic continuation at the "end" of this curve coincides with the derivative $f'$. Is it true that $f(z)=Ce^{z}$ or there may be some other such functions? (The original question in the book reads "What can be said about $f$?" :?)

If a function $f$ such that $f=f'$ on some curve,we can prove $f=Ce^{z}$ on that curve.
Then by the limit point theorem,There is some limit point such that $f=Ce^{z}$ on $\mathbb{C}$,then $f=Ce^{z}$?
Am I miss some thing?

[quote="statistical"]zhaobin,   he means a different problem.  It is not assumed that $f(z) = f'(z)$, but that $g(z) = f'(z)$ where (for $z$ close to 0)  $g$ is the result of analytic continuation of $f$ around the loop.[/quote]
Thank you.I get what am wrong. :blush:

If nontrivial solutions exist then the dimension of the space of (germs of) functions that can be obtained by monodromy from $f$ would be infinite. That sounds impossible, but I don't know enough complex analysis to be sure of that.  (If the space is finite-dimensional then $f$ satisfies a linear differential equation with constant coefficients, and the solutions of these are well-known to be entire and so the monodromy is trivial: $f = f'$.)

What do you mean by the "[b]space[/b] of germs that can be obtained from $f$ by monodromy"? The addition of two such germs does not result in such a germ again :?. If you just mean that these germs must be linearly dependent, that is not true: take any analytic function $F$ in the vertical strip $1/2<Re\,z<2$ such that the shifts $F(z-2\pi i k)$, $k\in\mathbb Z$ are linearly independent and put $f(z)=F(\log z)$.

[quote="fedja"]What do you mean by the "[b]space[/b] of germs that can be obtained from $f$ by monodromy"? [/quote]

The vector space generated by such germs.  But as you pointed out, these need not be finite dimensional; after posting I noticed examples such as $\exp( \log(z)^{3})$.


Find all functions $ f: \mathbb R\to\mathbb R$ such that
\[ f(x\plus{}f(y\minus{}x))\equal{}y\plus{}2f(x)\]
for all $ x,y\in\mathbb R$.

[quote="nayel"]Find all functions $ f: \mathbb R\to\mathbb R$ such that
\[ f(x \plus{} f(y \minus{} x)) \equal{} y \plus{} 2f(x)
\]
for all $ x,y\in\mathbb R$.[/quote]

$ x\equal{}0$ and $ y\equal{}\minus{}2f(0)$ implies $ f(f(\minus{}2f(0)))\equal{}0$

Then $ y\equal{}x\plus{}f(\minus{}2f(0))$ implies $ f(y\minus{}x)\equal{}0$ and so $ f(x)\equal{}x\plus{}f(\minus{}2f(0))\plus{}2f(x)$ and so $ f(x)\equal{}a\minus{}x$ for some real $ a$.

Plugging back this value in original equation, we get $ a\equal{}0$ and so the unique solution $ \boxed{f(x)\equal{}\minus{}x}$.

Wow... nice and quick solution. I didn't think it would be this easy...

hi  nayel!! if   f(0)=c   we have f(c)=0 and  f(f(c))=4c   then f(2c)=c    f(f(c))=f(2c)=c   so 4c=c  c=0

in the main equality x=y  then f(x)=-x  this problem is very obviouse u try to solve my problem ( IRAN!!! navid safaee)


geometry
perimeter
Pythagorean Theorem
puzzles

There is a right triangle with a perimeter of 5. Two of the sides have a length of two. How long is the third side?

Do you mean each of two sides has length 2?

i mean that two of the sides have a length of 2, the goal is to find the length of the other side

[hide]Well given that the perimeter is 5, the answer would be 1, but a right triangle can't exist with lengths 2 2 1. So the answer is that no such triangle can exist.[/hide]

Based on what has been presented in posts above,

there is NO SOLUTION to this problem.

Case 1a) rem's interpretation

hypotenuse of length 2 & one of the legs of length 2 -- not possible because it's a right triangle

Case 1b) rem's interpretation

each leg is of length 2 --> This makes the hypotenuse be
sqrt(8).  But the side lengths added up would be
2 + 2 + sqrt(8) > 5, which is the assigned perimeter.  Not possible.

Case 2a) "Two of the sides have a [total] length of two."

Without loss of generality, let the hypotenuse be 2-x, one of the legs be x, and the remaining leg be y.  As for the perimeter,
y + x + 2 - x = 5 leads to y + 2 = 5 leads to y = 3, but this is a
contradiction because y, a leg, is longer than 2-x, the hypotenuse.  Not possible.

Case 2b) "Two of the sides have a [total] length of two."

Without loss of generality, let one of the legs be x, the other
leg be 2-x, and the hypotenuse be y.  For the perimeter,
x + 2 - x + y = 5 leads to 2 + y = 5 leads to y = 3.  Now, use
the Pythagorean Theorem:  x^2 + (2-x)^2 = 3^2 leads to...
x^2 + 4 - 4x + x^2 = 9 leads to 2x^2 - 4x + 4 = 9 leads to ...
2x^2 - 4x - 5 = 0.  The positive root from this equation for x
doesn't work, because 1) when added to y = 3, makes the partial perimeter already too large, and 2) the quantity 2-x associated with it is negative.  Not possible.


Fie A1,A2,A3 un triunghi ascutitunghic si H ortocentru sau. Notam cu Hi piciorul inaltimii duse din Ai, iar cu ai,cu i=1,2,3 lungimea laturii opuse acestui varf.Sa se arate ca are loc relatia:

a1/HH1+a2/HH2+a3/HH3=2(a1/HA1+a2/HA2+a3/HA3)

Notam mai bine triunghiul cu $\ ABC$ cu $\ H$ ortocentru si $\ D,E,F$ picioarele inaltimilor din $\ A,B,C$.
$\ AH=2R\cos A$ si analoagele

$\ HD=2R\cos B\cos C$ si analoagele

$\cos^{2}\frac{A}{2}=\frac{1+\cos A}{2}$ si analoagele

$\cos^{2}\frac{A}{2}=\frac{p(p-a)}{bc}$ si analoagele

Combinand toate relatiile astea de mai sus si tinand cont de unele smecherii cu arii iese concluzia.

Eu nu am facut-o asa !!!Iar problema este frumoasa datorita solutiei care nu este deloc grea!!!hai mai incercati-o!!Gasiti si alte solutii!!

haideti!!nu mai da nimeni nicio solutie???

Pai ce, nu e de ajuns o solutie?E si asta frumoasa.Mai bine posteaza solutia ta. :wink:

$\frac{a1HA1}{HH1 HA1}+\frac{a2HA2}{HH2 HA2}+ \frac{a3 HA3}{HH3 HA3}= \frac{a1HA1+a2 HA2+a3HA3}{k}$
$= \frac{a1(H1A1-HH1)+a2(H2A2-HH2)+a3(H3A3-HH3)}{k}= \frac{a1H1A1+a2 H2A2+a3 H3A3 -(a1 HH1+a2HH2+a3HH3)}{k}=\frac{2S+2S+2S-2S}{k}=\frac{4S}{k}=2\frac{2S}{k}=2\frac{a1HH1+a2HH2+a3HH3}{k} =2\frac{a1HH1}{HH1HA1}+\frac{a2HH2}{HH2 HA2}+\frac{a3HH3}{HH3 HA3}
=2(frac{a1}{HA1}+\frac{a2}{HA2}+\frac{a3}{HA3}) $

mor de nervi!!!!!! M-am chinuit 30 de minute sa-l scriu asa ca scuze daca are greseli !!!Asa se intampla cand nu citesti latexul si inveti de pe forum!!

Aoleu ce e aici!! :) Nici nu pot sa citesc.Te-as sfatui sa inveti Latexul asta mai repede.

Problema a aparut in GM. nr.12/1998, este vorba de problema C:2108.
Are insa loc si urmatoarea analoaga spatiala:

Fie A1A2A3A4 un tetraedru ortocentric al carui ortocentru H se gaseste in interiorul tetraedrului.
Hotam cu Hi piciorul inaltimii duse din varful Ai, iar cu Si aria fetei opuse aceluiasi varf.
Sa se asrate ca are loc relatia:
S1/HH1+S2/HH2+S3/HH3+S4/HH4=3.(S1/HA1+S2/HA2+S3/HA3+S4/HA4).

Cred ca solutia problemei din GM a aparut si in GM(n-am verificat). Solutia celor 2 probleme
(a celei din GM si a analoagei spatiale) poate fi insa gasita in cartea noastra:
            Mihai Miculita, Dan Branzei: "ANALOGII TRIUNGHI-TETRAEDRU",
                          aparuta la Ed. PARALELA 45, Pitesti-2000.

[quote="Andrei Mihnea Stefan"][color=darkred]Fie $ A_{1}A_{2}A_{3}$ un triunghi ascutitunghic si $ H$ ortocentru sau. Pentru $ k\in\overline{1,3}$ notam $ H_{k}$- piciorul inaltimii duse din $ A_{k}$,
iar cu $ a_{k}$-  lungimea laturii opuse acestui varf. Sa se arate ca are loc relatia $ \sum_{k = 1}^{3}\frac{a_{k}}{HH_{k}}= 2\cdot\sum_{k = 1}^{3}\frac{a_{k}}{HA_{k}}$.[/color][/quote]

[color=darkblue][b]Demonstratia lui Andrei Mihnea[/b]. Se stie ca $ \{\begin{array}{c}HH_{1}\cdot HA_{1}= HH_{2}\cdot HA_{2}= HH_{3}\cdot HA_{3}= K\\ \ a_{1}\cdot H_{1}A_{1}= a_{2}\cdot H_{2}A_{2}= a_{3}\cdot H_{3}A_{3}= 2\cdot S\end{array}$, unde $ S$ este aria $ \triangle A_{1}A_{2}A_{3}$. Asadar,
$ \sum_{k = 1}^{3}\frac{a_{k}}{HH_{k}}=$ $ \sum_{k = 1}^{3}\frac{a_{k}\cdot HA_{k}}{HH_{k}\cdot HA_{k}}=$ $ \frac{1}{K}\cdot\sum_{k = 1}^{3}a_{k}\cdot HA_{k}=$ $ \frac{1}{K}\cdot\sum_{k = 1}^{3}a_{k}\cdot (H_{k}A_{k}-HH_{k}) =$ $ \frac{1}{K}\cdot(\sum_{k = 1}^{3}a_{k}\cdot H_{k}A_{k}-\sum_{k = 1}^{3}a_{k}\cdot HH_{k}) =$

$ \frac{3\cdot 2S-2S}{K}=$ $ \frac{4S}{K}=$ $ \frac{2}{K}\cdot 2S =$ $ \frac{2}{K}\cdot\sum_{k = 1}^{3}a_{k}\cdot HH_{k}=$ $ 2\cdot\sum_{k = 1}^{3}\frac{a_{k}\cdot HH_{k}}{HH_{k}\cdot HA_{k}}=$  $ 2\cdot\sum_{k = 1}^{3}\frac{a_{k}}{HA_{k}}$.

[b]Andrei Mihnea Stefan zice :[/b][/color] [color=darkred]"Mor de nervi !!!!!!
M-am chinuit 30 de minute sa-l scriu asa ca scuze daca are greseli !!!
Asa se intampla cand nu citesti latexul si inveti de pe forum !!".[/color]

[color=darkblue]Mare adevar grait-ai, [b]A.M.S.[/b] ([b]a[/b]mbitie, [b]m[/b]eticulozitate, [b]s[/b]arguinta) ! Este pacat sa nu redactezi in LaTeX o asemenea [u]frumoasa demonstratie[/u] !
Eu am invatat LaTeX - ul in cel mult trei saptamani cu un ochi in [b]LaTeX Help[/b] si cu celalalt pe [b]Forum[/b].
Sper ca in cel mult o luna sa-mi oferi placerea unui text scris frumos in LaTeX. Iti urez mult succes !

[b]P.S.[/b] Nu mi-au trebuit decat aprox. 10 minute pentru redactarea acestui text.[/color]


A mails a letter to B. One letter in n is lost in the mail, find the chance B received the letter.

[hide="hint"]If 1 letter in n is lost, then how many letters in n aren't lost?[/hide]
[hide="solution"]
Well, if 1 letter in n is lost, then $ n\minus{}1$ letters in n aren't lost, so the answer is $ \frac{n\minus{}1}{n}$[/hide]

[quote="Fraenkel"][quote="andersonw"][hide="hint"]If 1 letter in n is lost, then how many letters in n aren't lost?[/hide]
[hide="solution"]
Well, if 1 letter in n is lost, then $ n \minus{} 1$ letters in n aren't lost, so the answer is $ \frac {n \minus{} 1}{n}$[/hide][/quote]

out of the n letters only one belongs to B.[/quote]

True, but if you see words to the effect of "what is the [u]chance[/u] that such and such?"  you should read them as if they said, "what is the [u]probability[/u] that such and such?"

In this case there are only two possibilities: 1) the letter is lost, or 2) the letter is not lost. The letter is not lost if and only if the person to whom it is addressed, B, receives it. The probability that a letter, any letter, is not lost is: $ \frac{n \minus{} 1 \text{ not lost}}{n \text{ letters sent}}$. Therefore, the chance (probability) that B receives the letter sent to him is: $ \frac{n\minus{}1}{n}$.


Let be given a triangle $ABC.$ Prove that: $3\sum\cot{A}\ge\sum\cot{\frac{A}{2}}.$

[hide]
Let $x=\tan \frac{A}{2}$, $y=\tan \frac{B}{2}$, $z=\tan \frac{C}{2}$. After substituting, the inequality becomes
\[\frac{3}{2}\sum_\text{cyc}\frac{1-x^{2}}{x}\geq \sum_\text{cyc}\frac{1}{x}\]
After multiplying by 2 and cancelling, this is equivalent to
\[\sum_\text{cyc}\frac{1}{x}\geq 3\sum_\text{cyc}x\]
or, after clearing denominators,
\[xy+yz+zx\geq 3xyz(x+y+z)\]
But we have $z=\cot \left( \frac{B}{2}+\frac{C}{2}\right)=\frac{1-xy}{x+y}$, so $xy+yz+zx=1$. Homogenizing, we rewrite as
\[(xy+yz+zx)^{2}\geq 3xyz(x+y+z)\]
But this is evident from
\[(xy+yz+zx)^{2}-3xyz(x+y+z)=\frac{x^{2}(y-z)^{2}+y^{2}(z-x)^{2}+z^{2}(x-y)^{2}}{2}\geq 0\]
[/hide]

Nice [b]scorpius119[/b]. And here is my solution.
We have: $3\sum\cot A\ge\sum\cot\frac{A}{2}$
$\Longleftrightarrow 3\sum\frac{\cos{A}}{\sin{A}}\ge\sum\frac{p-a}{p}$
$\Longleftrightarrow 3\sum\frac{b^{2}+c^{2}-a^{2}}{2bc\sin A}\ge\frac{p}{r}$
$\Longleftrightarrow 3\frac{\sum a^{2}}{4S_{ABC}}\ge\frac{p}{r}$
$\Longleftrightarrow 3\sum a^{2}\ge\left(\sum a\right)^{2}$
$\Longleftrightarrow\sum (a-b)^{2}\ge 0$, what is truly.


Prove that the equation has an infinit number of solutions over R:

x^{x}=[x]; where x={x}+[x].

by Ion Nedelcu

Let us take n a natural number, n>1. Take the function f(x)=x^x-n-n defined for n<=x<=n+1. Of course, f is continuous and f(n)<0, f(n+1)>0. Thus we can find n<x<n+1 such that f(x)=0. Thus, x is a solution. Of course, for different n we obtain different solutions.


calculus
integration
logarithms
limit
search
calculus computations

I thinks its pretty easy if you express both as an infinite series and then take the Cauchy product. Integrate term by term and evaluate. It is just a little messy.

If we define $\text{dilog }(x)=\int_{1}^{x}\frac{\ln t}{1-t}\,dt$ then
\[\int_{0}^{1}\ln x\cdot \ln (1-x)\,dx=\left [(x\ln x)\ln (1-x)-x\ln x+2x+\text{dilog}(x)+\ln (x-1)\right ]_{x=0}^{1}. \]
Then just evaluate the expression.

Edit: I seems that this approach does not work. If $f(x)=(x\ln x)\ln (1-x)-x\ln x+2x+\text{dilog}(x)+\ln (x-1)$ then $\lim_{x\to 1-}f(x)=-\infty$. My mistake.

I know this has appeared in this forum, and I'm pretty sure I have posted an answer, but I can't figure out what key words or words to search for.

Another way to do it with series is noting that
$\ln(1-x)\ln x = \sum_{k=1}^\infty \frac{1}{k^{2}}-\sum_{k=1}^\infty \frac{x^{k}+(1-x)^{k}}{k^{2}}$


How many 3-digit numbers are there in which the digits are such

that if you create all the different 3 -digit numbers from them, and

you add thest numbers, the sum is divisible by 111?

[hide="Case 1"] Let's think about the numbers with three distinct digits $abc = 100a+10b+c$.  Each digit appears twice in each spot - twice as a hundreds digit, twice as a tens digit, twice as a ones digit, so our sum is

$100(2a+2b+2c)+10(2a+2b+2c)+1(2a+2b+2c) = 222(a+b+c)$

Which is, of course, divisible by $111$. [/hide] [hide="Case 2"] Let's think about the numbers with two distinct digits $abb$.  Now, $a$ appears once in each spot; the sum is

$100(a+2b)+10(a+2b)+1(a+2b) = 111(a+2b)$

Which is still divisible by $111$. [/hide] [hide="Case 3"] The numbers with only one distinct digit $aaa$ are, of course, divisible by $111$. [/hide] [hide="Making our final answer..."] Every $3$-digit number.  There are $\boxed{900}$ of them. [/hide]

I think you overcounted, t0rajir0u.. You can't take numbers having zero(s) as digit(s). That is, you can, but the formulae would be different.

Ah.  We can only form $3$-digit numbers.  In that case,

[hide="Case 4"] Numbers with one zero and two distinct digits $ab0$ form the permutations $ab0, a0b, ba0, b0a$ which sum to

$100(2a+2b)+10(a+b)+(a+b) = 211(a+b)$

Which is never divisible by $111$. [/hide] [hide="Case 5"] Numbers with one zero and two of the same digits $aa0$ form the permutations $aa0, a0a$ which sum to

$100(2a)+10(a)+a = 211a$

Which is also never divisible by $111$. [/hide] [hide="Case 6"] Numbers with two zeroes $a00$ form only that permutation and are never divisible by $111$. [/hide] [hide="Hence our, ahem, real answer..."] Is the number of $3$-digit numbers containing no zeroes.  This is $9 \cdot 9 \cdot 9 = \boxed{729}$. [/hide]


John is walking east at a speed of 3 miles per hour, while Bob is also walking east, but at a speed of 5 miles per hour.  If Bob is now 1 mile west of John, how many minutes will it take for Bob to catch up to John?

$ \textbf{(A) } 30 \qquad \textbf{(B) } 50 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 120$

[hide="Solution"]
$(A)$

Bob is catching up to John at a rate of $5-3=2$ mph.  Using $d=rt$ we find that $1 \text{mile}=\left(2\frac{\text{miles}}{\text{hour}}\right)t\Rightarrow t=1/2 \text{hour}$.  This is just 30 minutes.[/hide]

[color=indigo]another way to do it:
[hide]
say that john starts at b and bob starts at a:
_____B_____^_____Z
A____*_____*_____Z

^=where john is after 30 min (if B-Z is three miles)
* =where bob is after 15 min and the second * where bob is after 30min, duh! (if A-Z=5miles)  so we see that after 30min john and bob meet.[/hide][/color]


Let ABC be a triangle with incenter I. E,F be the points on AB,BC, respectively, such that BE=CF.
The ray BF and CE intersect at M. Let $ A_{1},A_{2},A_{3}$ be the midpoints of BC,EF,IM, respectively.Prove that: $ A_{1},A_{2},A_{3}$ are collinears.

Are you sure?maybe you mean $ AF$ and $ CE$ intersects at $ M$?

[quote="NguyenDungTN"]Let $ ABC$ be a triangle with incenter $ I$. $ E$, $ F$ be the points on $ AB$, $ BC$, respectively, such that $ BE \equal{} CF$. The ray $ BF$ and $ CE$ intersect at $ M$. Let $ A_{1}$, $ A_{2}$, $ A_{3}$ be the midpoints of $ BC$, $ EF$, $ IM$, respectively. Prove that: $ A_{1}$, $ A_{2}$, $ A_{3}$ are collinears.[/quote]
I think it's equivalent to http://www.mathlinks.ro/Forum/viewtopic.php?t=168094  :wink:
And can we discuss it now?


A graduate of Washington School earns her living by planning meals and buying groceries for ladies who do not like to do it themselves.  From one family, she receives $ 8.00 a month plus 1/3 of every dollar less than$80 per month that the family has to spend for food.  How much does this family pay her for a month when the food costs them $ 62.00?      (Round to the nearest cent.)
A.$6.00	B.$ 14.00  C.$16.33	D.$ 28.67  E.$34.67

[hide][hide=solution]First she recieves $8.00/month
Total:$8.00
Then she recieves 1/3 of the difference between what the family spent on groceries and $80.
In this problem the difference is $18 ($80-$62).  $18/3 = $6
$8+$6=$14
Therefore the answer choice is [b]B) $14.00[/b][/hide][/hide]

Zukex, can yo please show your solution? Thanks

[hide="my solution"] The graduate recieves 8.00 a month
She gets 1/3 of the amount that her family spent on groceries and the 80 bucks.
8+6 dollars = 14 dollars.
I'm not putting in Latex symbols, sorry :(
So the answer would be B[/hide]