floor function
number theory
equation
Summation
IMO
IMO 1968

Let $n$ be a natural number. Prove that \[ \left\lfloor \frac{n+2^0}{2^1} \right\rfloor + \left\lfloor \frac{n+2^1}{2^2} \right\rfloor +\cdots +\left\lfloor \frac{n+2^{n-1}}{2^n}\right\rfloor =n. \]
[hide="Remark"]For any real number $x$, the number $\lfloor x \rfloor$ represents the largest integer smaller or equal with $x$.[/hide]

It immediately comes from base 2 representation.
Indeed, suppose $n=\overline{a_{n}...a_2a_1a_0}$ (actually, we know that it containes at most $n-1$ digit for $n\geq 1$). And we have $\left[\frac{n+2^{k-1}}{2^{k}}\right]=\left[\frac{n}{2^{k}}+\frac{1}{2}\right]=\overline{a_{n-1}...a_{k}}+a_{k-1}$ :!:. Therefore the whole sum is $\sum a_k(2^k-1)+\sum a_{k-1}= \sum a_k2^k=n$.

We can use telescoping. [hide="Hint"]For any real number $ \alpha$, we have $ \left\lfloor \alpha \plus{} \frac12 \right\rfloor \equal{} \lfloor 2\alpha\rfloor \minus{} \lfloor \alpha \rfloor$.[/hide]

[quote="keira_khtn"]Let $ n$ be a natural number. Prove that
\[ \left\lfloor \frac {n \plus{} 2^0}{2^1} \right\rfloor \plus{} \left\lfloor \frac {n \plus{} 2^1}{2^2} \right\rfloor \plus{} \cdots \plus{} \left\lfloor \frac {n \plus{} 2^{n \minus{} 1}}{2^n}\right\rfloor \equal{} n.
\]
[hide="Remark"]For any real number $ x$, the number $ \lfloor x \rfloor$ represents the largest integer smaller or equal with $ x$.[/hide][/quote]
I'll complete Johan Gunardi's hint.

First a lemma:
$ \left\lfloor \alpha \plus{} \frac{1}{2} \right\rfloor \equal{} \left\lfloor 2\alpha \right\rfloor \minus{} \left\lfloor \alpha \right\rfloor$
Let $ \alpha \equal{} n \plus{} \delta$ where $ n \in \mathbb{Z}, \delta \in [0;1)$.
Then it's equivalent to proving: $ \left\lfloor \delta \plus{} \frac{1}{2} \right\rfloor \plus{} \left\lfloor \delta \right\rfloor \equal{} \left\lfloor 2\delta\right\rfloor$ $ \iff$ $ \left\lfloor \delta \plus{} \frac{1}{2} \right\rfloor \equal{} \left\lfloor 2\delta\right\rfloor$. (Which is true by taking the two cases $ \delta < \frac{1}{2}$ and $ \delta \ge \frac{1}{2}$)

We want to prove that:
$ \sum_{k\equal{}1}^n\left\lfloor \frac {n}{2^k}\plus{}\frac{1}{2} \right\rfloor \equal{} n$
But by the lemma:
$ \sum_{k\equal{}1}^n\left\lfloor \frac {n}{2^k}\plus{}\frac{1}{2} \right\rfloor \equal{} \sum_{k\equal{}1}^n\left\lfloor \frac {n}{2^{k\minus{}1}} \right\rfloor \minus{} \left\lfloor \frac {n}{2^k} \right\rfloor \equal{}  \left\lfloor n \right\rfloor \minus{} \left\lfloor \frac {n}{2^n} \right\rfloor \equal{} n$ so it is obviously true :)

[quote="manjil"]You can also use Hermite's Identity to prove this![/quote]
Could you explain a bit further?

By Hermite's Identity, we have $ \lfloor x \rfloor \plus{} \left\lfloor x\plus{}\frac12 \right\rfloor \equal{} \lfloor 2x \rfloor$. Hence $ \left\lfloor x\plus{}\frac12 \right\rfloor\equal{}\lfloor 2x \rfloor\minus{} \lfloor x \rfloor$.

[quote="cosinator"]Here is a rather contrived proof (but still a proof nonetheless):

Consider the binary representation of $ n$: $ n \equal{} 2^{u_0} \plus{} 2^{u_1} \plus{} \cdots \plus{} 2^{u_t}$.  Then it is easy to see that

$ \left\lfloor\frac {n \plus{} 2^{k \minus{} 1}}{2^k}\right\rfloor \equal{} 2^{u_i \minus{} k} \plus{} 2^{u_{i \plus{} 1} \minus{} k} \plus{} \cdots \plus{} 2^{u_t \minus{} k} \plus{} (0)\text{ if }u_{i \minus{} 1} < k \minus{} 1\text{ and } \plus{} (1)\text{ if }u_{i \minus{} 1} \equal{} k \minus{} 1$

for $ u_i\ge k$.  Thus, the sum in question is equivalent to

$ (2^{u_0 \minus{} 1} \plus{} 2^{u_0 \minus{} 2} \plus{} \cdots \plus{} 1) \plus{} (2^{u_1 \minus{} 1} \plus{} 2^{u_1 \minus{} 2} \plus{} \cdots \plus{} 1) \plus{} \cdots \plus{} (2^{u_t \minus{} 1} \plus{} 2^{u_t \minus{} 2} \plus{} \cdots \plus{} 1) \plus{} t \plus{} 1$

$ \equal{} 2^{u_0} \minus{} 1 \plus{} 2^{u_1} \minus{} 1 \plus{} \cdots \plus{} 2^{u_t} \minus{} 1 \plus{} t \plus{} 1$

$ \equal{} n$[/quote]

The given sum is equal to the sum $ \sum \left\lfloor\frac {n}{2^k}\right\rfloor +$ sum of digits of $n$ in binary.

If we compare $ \sum \frac {n}{2^k} - \sum \left\lfloor\frac {n}{2^k}\right\rfloor $ it is clear the error equals the sum of these digits. If the digit represents $2^x$ then its contribution to the error is going to be $\frac{1}{2}$ for $k=x+1$,$\frac{1}{4}$ for $k=x+2$ and so on.

Can somebody check if this solution works?

[hide="Induction"]
I induct to prove that this holds for all $n$, the base case is trivial. Assume $n$ is even, let $n=2n'$.

$\sum_{k=0}^{\infty}\left\lfloor\frac{2n'}{2^{k+1}}+\frac{1}{2}\right\rfloor = \sum_{k=0}^{\infty}\left\lfloor\frac{n'}{2^k}+\frac{1}{2}\right\rfloor = \sum_{k=-1}^{\infty}\left\lfloor\frac{n'}{2^{k+1}}+\frac{1}{2}\right\rfloor=\left(\sum_{k=0}^{\infty}\left\lfloor\frac{n'}{2^{k+1}}+\frac{1}{2}\right\rfloor\right)+\left\lfloor n'+\frac{1}{2}\right\rfloor$, since $n'$ is an integer $<n$, by induction hypothesis we have it equaling $2n'=n$, so done in this case.

Assume $n$ is odd, let $n=2n'-1$.
$\sum_{k=0}^{\infty}\left\lfloor\frac{2n'-1}{2^{k+1}}+\frac{1}{2}\right\rfloor=\sum_{k=0}^{\infty}\left\lfloor\frac{n'}{2^{k}}+\frac{1}{2}-\frac{1}{2^{k+1}}\right\rfloor = \sum_{k=0}^{\infty}\left\lfloor\frac{2n'+2^k}{2^{k+1}}-\frac{1}{2^{k+1}}\right\rfloor$. Note that $\left\lfloor\frac{2n'+2^k}{2^{k+1}}-\frac{1}{2^{k+1}}\right\rfloor \neq \left\lfloor\frac{2n'+2^k}{2^{k+1}}\right\rfloor$ iff $2^{k+1}\mid 2n'+2^k$ or $v_2(2n'+2^k)\ge k+1$, since $v_p(a+b)=v_p(a)$, given $v_p(b)>v_p(a)$, we must have $v_2(2n')=k$. Thus our sum is $\sum_{k=0}^{\infty}\left\lfloor\frac{2n'+2^k}{2^{k+1}}\right\rfloor - 1=\sum_{k=0}^{\infty}\left\lfloor\frac{n'+2^{k-1}}{2^{k}}\right\rfloor - 1=\sum_{k=-1}^{\infty}\left\lfloor\frac{n'+2^{k}}{2^{k+1}}\right\rfloor - 1=\sum_{k=0}^{\infty}\left\lfloor\frac{n'+2^{k}}{2^{k+1}}\right\rfloor+\left\lfloor n'+\frac{1}{2}\right\rfloor - 1$, which by induction hypothesis is $2n'-1=n$, so we are done$\blacksquare$
[/hide]

I think yours works haha since I decided to use induction as well.

[hide=Solution]Let $n = 2^k + m$ where $0 \leq m < 2^k$. We will instead prove $$\lfloor \frac{2^k+m+1}{2} \rfloor + \lfloor \frac{2^k + m + 2}{4} \rfloor + \cdots + \lfloor \frac{2^k + m + 2^k}{2^{k+1}} \rfloor = 2^k + m.$$ The next denominator would be too big.

Now, we use strong induction on $k$. The base case, $k = 0$ works clearly since $$\lfloor \frac{2+m}{2} \rfloor = 1 = 2^0 + 0.$$ Now, assume $k-1$ works. For $k + 1$, notice that the left hand side expression is equal to $$(2^{k-1} + 2^{k-2} + \cdots + 2 + 1 + 1) + \lfloor \frac{m+1}{2} \rfloor + \cdots + \lfloor \frac{m+2^{k-1}}{2^k} \rfloor + 0 .$$ By the inductive hypothesis (since $m < 2^k$) the expression is equal to $$2^k + m.$$ As desired. [/hide]

Quite nice.
-----
The core of this problem is in the following:

[b][color=#f00]Lemma:[/color][/b] For positive, real $n$, we have $$\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n+1}{2}\right\rfloor=\left\lfloor{n}\right\rfloor.$$
[i]Proof:[/i] Assume $\lfloor{n}\rfloor$ is even (the case for odd is similar). We have $n=2m+\text{fractional n}$, with $m$ an integer. Thus, we get $\left\lfloor\frac{n}{2}\right\rfloor=m$ after some computation. $$n+1=2m+1+\text{fractional n} \rightarrow\left\lfloor\frac{n+1}{2}\right\rfloor=m+\left\lfloor\frac{1+\text{\text{fractional} n}}{2}\right\rfloor=m.$$ The required result follows. $\square$

Our problem now reduces as $\sum_{k=1}^n \left\lfloor{n}\right\rfloor-\left\lfloor\frac{n}{2^n}\right\rfloor=n$, as needed. $\blacksquare$

Note that $[x]+[x+1/2]=[2x]$ for all $x>0.$ This follows from casework on fractional part. Then LHS is $[n]-[n/2^n]=n.$ Q.E.D.

Here is a better version of superagh's solution. (In my opinion, his induction hypothesis application does not seem valid enough, as you cannot mix $k$ and $m$ just like that, as the initial statement has only one variable and not two.)

We use strong induction on $n$. For $n=1$ we have only one summand and it is $\left\lfloor \frac{1+2^0}{2^1}\right \rfloor = 1$. Now assume the required equality holds for all positive integers less than $n$ and write $n=2^k+m$ where $0 \leq m \leq 2^{k}-1$. Since $\left\lfloor \frac{2^k + m + 2^s}{2^{s+1}} \right\rfloor = 2^{k-s-1} + \left\lfloor \frac{m+2^s}{2^{s+1}} \right\rfloor$ for $s\leq k-1$ and $\left\lfloor \frac{2^k + m + 2^k}{2^{k+1}} \right \rfloor= 1$, on the left we have $\left\lfloor \frac{m+2^0}{2^1} \right\rfloor + \left\lfloor \frac{m+2^1}{2^2} \right\rfloor + \left\lfloor \frac{m+2^2}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{m+2^{n-1}}{2^n} \right\rfloor + 2^{k-1} + 2^{k-2} + \cdots + 1 + 1 = \left\lfloor \frac{m+2^0}{2^1} \right\rfloor + \left\lfloor \frac{m+2^1}{2^2} \right\rfloor + \left\lfloor \frac{m+2^2}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{m+2^{n-1}}{2^n} \right\rfloor + 2^k$, while on the right we have $2^k + m$. Since $\left\lfloor \frac{m+2^s}{2^{s+1}} \right\rfloor = 0$ for $s\geq k$ (as $m < 2^k$), we reduce to proving
        $$ \left\lfloor \frac{m+2^0}{2^1} \right\rfloor + \left\lfloor \frac{m+2^1}{2^2} \right\rfloor + \left\lfloor \frac{m+2^2}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{m+2^{k-1}}{2^k} \right\rfloor = m. $$
        For if $m \geq k$, then on the left we can add the $0$ summands $\left\lfloor \frac{m+2^{s-1}}{2^s}\right \rfloor$, $s=k+1,\ldots,m$ and to desire an equality which follows from the induction hypothesis as $m < 2^k < n$.
        Now suppose $m\leq k-1$. Then $\left\lfloor \frac{m + 2^{s-1}}{2^s} \right \rfloor = \left \lfloor \frac{m}{2^s} + \frac{1}{2} \right \rfloor = 0$ for $s\geq m+1$ since $\frac{m}{2^{m+1}} < \frac{1}{2} \Leftrightarrow 2^m > m$ (by a simple induction on $m$). So we may remove the $0$ summands $\left\lfloor \frac{m+2^{s-1}}{2^s}\right \rfloor$, $s=m+1,\ldots,k-1$ and to desire an equality which follows from the induction hypothesis as $m < 2^k < n$.

Simplify the problem into the following: \[\left \lfloor \frac{n}{2} + \frac{1}{2} \right \rfloor + \left \lfloor \frac{n}{4} + \frac{1}{2} \right \rfloor + \left \lfloor \frac{n}{8} + \frac{1}{2} \right \rfloor + \left \lfloor \frac{n}{16} + \frac{1}{2} \right \rfloor + \cdots.\]
Now we use the identity that $\lfloor x \rfloor + \lfloor x + 1/2 \rfloor = \lfloor 2x \rfloor$, and rearranging we find $\lfloor x + 1/2 \rfloor = \lfloor 2x \rfloor - \lfloor x \rfloor$. Hence our initial sum becomes the following: \[\left(\lfloor n \rfloor - \left \lfloor \frac{n}{2} \right \rfloor \right)+ \left(\left \lfloor \frac{n}{2} \right \rfloor - \left \lfloor \frac{n}{4} \right \rfloor \right) + \left(\left \lfloor \frac{n}{4} \right \rfloor - \left \lfloor \frac{n}{8} \right \rfloor \right) + \cdots = \lfloor n \rfloor = n, \] as desired. $\blacksquare$


Find all $x,y \in N^{*}$ and $p$ is a prime number such that $x^{2}+p^{2}=y^{3}$

Strange problem. For example: $p=2699, x=26910, y=901$ works...

Yeah, that's exactly what I was talking about.

A result
Give $a,b,c \in N : a^{n}=b^{2}+c^{2}$
Prove that $a=x^{2}+y^{2}$


algebra
polynomial
calculus
derivative
algebra unsolved

P(x) is  a polynomial which is positive (and not 0) for all real numbers of x.prove that its derivative has only one root.

[quote="amcavoy"]Any polynomial of degree $n$ has $n-1$ possible critical points.  Therefore as ZetaX pointed out, you could have a polynomial of degree $4$, or $6$, or any even number greater than $2$ for which $f\left(x\right)\neq 0$ and $f'\left(x\right)=0$ has $>1$ solutions.[/quote]
Can be 1, for example $f(x)=x^{2n}+1$ or $x^{2n}+x^{2}+1$.


function
limit
inequalities
calculus
integration
triangle inequality
calculus computations

Suppose that f : R^n -> R^m is differentiable on an open conected set S and assume that there exist constants k>1 and C>0 such that for all x,y in R^n,
|| f(x) - f(y)|| is less than or equal to C ( ||x-y||^k).
Prove that f is constant on S.

Since $ f$ is differentiable on $ S$, for each $ x\in S$ there is a linear transformation $ D(x): \mathbb{R}^n\to \mathbb{R}^m$ so that,
\[ \lim_{\left \Vert h \right \Vert \to 0}\frac {\left\Vert f(x \plus{} h) \minus{} f(x) \minus{} D(x)h\right\Vert}{\left \Vert h \right \Vert} \equal{} 0
\]
Since for each $ x,h\in S$ we also have $ \frac {\left \Vert f(x \plus{} h) \minus{} f(x)\right \Vert}{\left \Vert h\right \Vert} \leq C\left \Vert h\right \Vert ^{k \minus{} 1}$, this implies for each $ x\in S$ that $ \lim_{\left \Vert h \right \Vert \to 0}\frac {\left \Vert f(x \plus{} h) \minus{} f(x)\right\Vert}{\left \Vert h\right \Vert} \equal{} 0$, and hence by the triangle inequality $ \lim_{\left \Vert h \right \Vert \to 0}\frac {\left \Vert D(x)h\right \Vert}{\left\Vert h \right \Vert} \equal{} 0$.  In particular, this implies $ \limsup_{\left\Vert h \right\Vert \to 0}\frac {\left \Vert D(x)h\right \Vert}{\left\Vert h \right \Vert} \equal{} 0$, and therefore by the linearity of $ D(x)$,
\[ \sup_{\left\Vert y\right \Vert \equal{} 1}\frac {\left \Vert D(x)y\right\Vert}{\left\Vert y\right \Vert} \equal{} \left \Vert D(x)\right \Vert \equal{} 0
\]
Since $ \left \Vert D(x)\right \Vert \equal{} 0$, we must have $ D(x) \equal{} \bf{0}_{m\times n}$ for each $ x\in S$.

Every open connected set is path connected, so for each $ x,y\in S$ there is a curve $ \gamma: [0,1]\to S$, $ \gamma(t) \equal{} \left( \begin{array}{c} \gamma_1(t) \\
\gamma_2(t) \\
\vdots \\
\gamma_n(t) \end{array} \right)$ with $ \gamma_k: [0,1]\to \mathbb{R}$ differentiable for each $ k$ and $ \gamma (0)\equal{}x,\gamma(1)\equal{}y$.  Then by the fundamental theorem of calculus,
\[ f(y) \equal{} f(x) \plus{} \int_0^1 D(\gamma (t))\gamma '(t)dt \equal{} f(x)
\]
where $ \gamma'(t) \equal{} \left( \begin{array}{c} \gamma_1'(t) \\
\gamma_2'(t) \\
\vdots \\
\gamma_n'(t) \end{array} \right)$, and hence $ f$ is constant.


Let $ a;b;c > 0$ and $ a\plus{}b\plus{}c \equal{}3$ . Prove that :

$ \frac{a}{\sqrt[3]{b^2\plus{}c^3}}\plus{}\frac{b}{\sqrt[3]{c^2\plus{}a^3}}\plus{}\frac{c}{\sqrt[3]{a^2\plus{}b^3}} \ge\ \frac{3}{\sqrt[3]{2}}$

Not true.  Take $ a \equal{} b \equal{} \frac {3}{2}$, $ c \equal{} 0$, then the left hand side is $ \sqrt [3]{\frac {3}{2}} \plus{} 1 < \frac {3}{\sqrt [3]{2}}$.  By continuity we can find an $ \epsilon > 0$ so that $ a \equal{} b \equal{} \frac {3 \minus{} \epsilon}{2}$, $ c \equal{} \epsilon$ for which the inequality doesn't hold.


limit
trigonometry
function
calculus
derivative
ratio
analytic geometry

Find \[ \lim_{x\to 0} \frac{\sin(\tan x)\minus{}\tan(\sin x)}{\arcsin(\arctan x) \minus{} \arctan(\arcsin x)}\]

While it can probably be decomposed somehow, whenever I see such a problem, I always jump to L'Hopital's Rule and you should too.

:o
I think we muts use it more than 4 times....
Maybe there is another way?

An [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=101678]old thread[/url] on this problem. The first thread is unfortunately lost forever.

So how did Huygens and Barrow find this limit "instantly from geometric considerations" (V.I. Arnold)?

I don't think they did. Don't trust anything Arnold says about the difficulty of problems.

Usually I'm quite sceptical about Arnold's claims too but here he does have a point. Let's generalize a bit. Suppose that $ f$ and $ g$ are analytic functions that have the same derivative $ a>0$ at $ 0$ but don't concide identically (in our case $ f(x)\equal{}\sin(\tan x)$ and $ g(x)\equal{}\tan(\sin x)$). Find
\[ \lim_{x\to 0}\frac{f(x)\minus{}g(x)}{f^{\minus{}1}(x)\minus{}g^{\minus{}1}(x)}\]
;)

Let $ f(x)\equal{}a_{1}x\plus{}a_{2}x^{2}\plus{}a_{3}x^{3}\plus{}\cdots\plus{}$

and $ g(x)\equal{}a_{1}x\plus{}b_{2}x^{2}\plus{}b_{3}x^{3}\plus{}\cdots$,

where $ a_{1}\neq 0$.

Then, simple calculations show that

$ f^{\minus{}1}(x)\equal{}\frac{1}{a_{1}}x\minus{}\frac{a_{2}}{a_{1}^{3}}x^{2}\plus{}\cdots\plus{}$

$ g^{\minus{}1}(x)\equal{}\frac{1}{a_{1}}x\minus{}\frac{b_{2}}{a_{1}^{3}}x^{2}\plus{}\cdots$.

It follows that $ \lim\limits_{x\rightarrow 0}\frac{f(x)\minus{}g(x)}{f^{\minus{}1}(x)\minus{}g^{\minus{}1}(x)}\equal{}\minus{}a_{1}^{3}$.

I hope I made no mistakes.

There's a possibility that $ a_2\equal{}b_2$, but I think it can be managed similarly.

@didilica How do you know that the second (third, fourth, etc.) order terms do not cancel entirely? ;) It is true that somewhere in the line you'll have non-equal coefficients but by that time the formulae for them in terms of the coefficients of $ f$ and $ g$ get so complicated that they'll be almost completely useless.

@mlok Similarly? Try it if $ a_i\equal{}b_i$ up to $ i\equal{}100$ and $ a_{101}\ne b_{101}$ :P

fedja,
You are right, I simply considered the very simple case when the first Taylor coefficients of $ f$ and $ g$ , namely $ a_{1}$ and $ b_{1}$, are equal. I did not know that you were thinking of such a general case.

Yes, I was :). To be completely precise, I should say (in addition to the obviously missing assumption $ f(0)\equal{}g(0)\equal{}0$) that the common value of the derivative is $ a\ne 0$ and that the functions $ f$ and $ g$ coincide up to order $ d$ but $ a_{d\plus{}1}\ne b_{d\plus{}1}$ (the answer actually depends on both $ a$ and $ d$ but on nothing else; moreover, when $ a\equal{}1$, the dependence on $ d$ disappears and that is what allowed Arnold to make his claim).

Let's consider the case $ a \equal{} 1$. Since $ f$ and $ g$ are invertible near $ 0$, we can let $ \phi \equal{} f\circ g^{ \minus{} 1}$. Then
$ \lim_{x\to 0}\frac {f(x) \minus{} g(x)}{f^{ \minus{} 1}(x) \minus{} g^{ \minus{} 1}(x)} \equal{} \lim_{x\to 0}\frac {\phi(x) \minus{} x}{\phi^{ \minus{} 1}(x) \minus{} x}$. If $ \phi(x) \equal{} x \plus{} c_nx^n \plus{} \dots$, then $ \phi^{ \minus{} 1}(x) \equal{} x \minus{} c_nx^n \plus{} \dots$, hence the limit is $ \minus{} 1$.

... this almost works, except that $ \phi^{\minus{}1}\equal{}g\circ f^{\minus{}1}$, and not $ f^{\minus{}1}\circ g$ as I wanted.  :ninja:

Erm... The whole point is to find the answer [b]without[/b] computing the Taylor coefficients of the inverse functions. Besides, as I said earlier, if you insist on the possibility of such computations, just do it for $ k\equal{}100$. Also, I didn't mean that all coefficients are $ a$, just that $ a_i\equal{}b_i$ for several first $ i$ (but $ a_2\ne a_1$, generally speaking). Are you paying attention Extremal, or you are back to your old habit to post whatever comes into your head first and think what it is that you posted afterwards? :roll:

[quote="fedja"]Besides, as I said earlier, if you insist on the possibility of such computations, just do it for $ k \equal{} 100$. Also, I didn't mean that all coefficients are $ a$, just that $ a_i \equal{} b_i$ for several first $ i$ (but $ a_2\ne a_1$, generally speaking).  :roll:[/quote]
I have corrected it up to your remark   :wink:
and I am sure that  ( $ M(a) \equal{} ( \minus{} 1)^{k \plus{} 1} \frac {1}{a^{k \plus{} 2}}$ or $ M(a) \equal{}  \minus{}  \frac {1}{a^{k \plus{} 2}}$)  it's easly seen : from how we find inverse function we must look at first coeficient.
I need few minute to check it . Wait..

The key is the geometric interpretation; the numerator is the vertical distance between the two curves and the denominator is the negative of the horizontal distance. If $ a\equal{}1$, these two distances are measured at almost the same point, and the ratio of the two distances is approximately the common slope $ 1$. With the negative, our answer is $ \minus{}1$.

I'd still call "instantly" an exaggeration. It doesn't take up much space on the page, but verifying it is tricky.

I think $ M(a) \equal{} \minus{} \frac {1}{a^{m}}$  (edit)
jmerry: in our case distance is equal to $ 0$ (say $ f(0) \equal{} g(0) \equal{} 0$) then geometric Reasons  don't work's for me  :( .

This is a [i]limit[/i]; it's not about what happens [i]at[/i] the point.

[quote="Extremal"]$ f \equal{} ax \plus{} a_{2}x^{2} \plus{} ... \plus{} a_{k} x^{k} \plus{} a_{k \plus{} 1}x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2})$
$ g \equal{} ax \plus{} a_{2}x^{2} \plus{} ... \plus{} a_{k} x^{k} \plus{} b_{k \plus{} 1}x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2})$
then we can find
$ f^{ \minus{} 1} \equal{} \alpha_{1}x \plus{} \alpha_{2}x^{2} \plus{} ... \plus{} \alpha_{k}x^{k} \plus{} a_{k \plus{} 1}M(a)x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2})$
and
$ g^{ \minus{} 1} \equal{} \alpha_{1}x \plus{} \alpha_{2}x^{2} \plus{} ... \plus{} \alpha_{k}x^{k} \plus{} b_{k \plus{} 1}M(a)x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2})$
after some calculation we can find $ M()$ and limit will equal to $ \frac {1}{M(a) }$
I think that $ M(a) \equal{} \minus{} \frac {1}{a^{k \plus{} 2}}$[/quote]
I want to correct some my mistakes :
in finally I get that
$ f^{ \minus{} 1} \equal{} \alpha_{1}x \plus{} \alpha_{2}x^{2} \plus{} ... \plus{} \alpha_{k}x^{k} \plus{} N_{1}(a,a_{2},...)x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2})$
and
$ g^{ \minus{} 1} \equal{} \alpha_{1}x \plus{} \alpha_{2}x^{2} \plus{} ... \plus{} \alpha_{k}x^{k} \plus{} N_{2}(a,a_{2},...)x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2})$
and
$ \lim... \equal{} \frac {a_{k \plus{} 1} \minus{} b_{k \plus{} 1}}{N_{1} \minus{} N_{2}} \equal{} \frac {1}{M(a)}$ where $ M(a)$ depends only $ a$
here is my solution:
when we consider
$ f \equal{} ax \plus{} a_{2}x^{2} \plus{} ... \plus{} a_{k} x^{k} \plus{} a_{k \plus{} 1}x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2})$
and try to find inverse function I seen that  inverse function has a form:
$ f^{ \minus{} 1} \equal{} M_{1}(a)x \plus{} M_{2}(a,a_{2})x^{2} \plus{} M_{3}(a,a_{2},a_{3})x^{3} \plus{} ... \plus{} M_{k}(a,a_{2},..a_{k})x^{k} \plus{} C*x^{k \plus{} 1}$ and how we find $ C$
(and also ffor $ g$ if $ g \equal{} ax \plus{} a_{2}x^{2} \plus{} ... \plus{} a_{k} x^{k} \plus{} b_{k \plus{} 1}x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2})$
then
$ g^{ \minus{} 1} \equal{} M_{1}(a)x \plus{} M_{2}(a,a_{2})x^{2} \plus{} M_{3}(a,a_{2},a_{3})x^{3} \plus{} ... \plus{} M_{k}(a,a_{2},..a_{k})x^{k} \plus{} B*x^{k \plus{} 1}$)
? we must substitute this expression into :
$ f \equal{} ax \plus{} a_{2}x^{2} \plus{} ... \plus{} a_{k} x^{k} \plus{} a_{k \plus{} 1}x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2})$
and we will get :
at first we must look on power $ (k \plus{} 1)$ all coeficient for $ x^{j}$ is zero for $ j \equal{} 1,k$ for $ j \equal{} k \plus{} 1$ also must be zero hence:

if we write all coeficient for $ k \plus{} 1$ we will get :
for $ f$ we have

$ aC \plus{} G_{1}(a,a_{1}) \plus{} G_{2}(a,a_{2},a_{3}) \plus{} ... \plus{} a_{k \plus{} 1}M^{k \plus{} 1}_{a} \equal{} 0$ remember last member !! it's important !!
and also for $ g$ we have
($ aB \plus{} G_{1}(a,a_{1}) \plus{} G_{2}(a,a_{2},a_{3}) \plus{} ... \plus{} b_{k \plus{} 1}M^{k \plus{} 1}_{a} \equal{} 0$ I ! note ! that $ G_{1},...,G_{k}$ is same  )
so $ C \equal{} \frac { \minus{} (G_{1}(a,a_{1}) \plus{} G_{2}(a,a_{2},a_{3}) \plus{} ... \plus{} a_{k \plus{} 1}M^{k \plus{} 1}_{a})}{a}$
and
for $ g$ we have
$ B \equal{} \frac { \minus{} (G_{1}(a,a_{1}) \plus{} G_{2}(a,a_{2},a_{3}) \plus{} ... \plus{} b_{k \plus{} 1}M^{k \plus{} 1}_{a})}{a}$
and to compute limit we have
$ \lim ... \equal{} \frac {f \minus{} g}{f^{ \minus{} 1} \minus{} g^{ \minus{} 1}}$ =$ \frac {ax \plus{} ... \plus{} a_{k \plus{} 1}x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2}) \minus{} (ax \plus{} ... \plus{} b_{k \plus{} 1}x^{k \plus{} 1} \plus{} O(x^{k \plus{} 2}))}{M_{1}(a)x \plus{} ... \plus{} M_{k}(a,a_{2},..a_{k})x^{k} \plus{} C*x^{k \plus{} 1} \minus{} (M_{1}(a)x \plus{} ... \plus{} M_{k}(a,a_{2},..a_{k})x^{k} \plus{} B*x^{k \plus{} 1})}$
$ \equal{} \frac {a_{k \plus{} 1} \minus{} b_{k \plus{} 1}}{C \minus{} B} \equal{} \frac {a_{k \plus{} 1} \minus{} b_{k \plus{} 1}}{\frac { \minus{} (G_{1}(a,a_{1}) \plus{} G_{2}(a,a_{2},a_{3}) \plus{} ... \plus{} a_{k \plus{} 1}M^{k \plus{} 1}_{a})}{a} \minus{} (\frac { \minus{} (G_{1}(a,a_{1}) \plus{} G_{2}(a,a_{2},a_{3}) \plus{} ... \plus{} b_{k \plus{} 1}M^{k \plus{} 1}_{a})}{a})}$=
$ \equal{} \frac {a_{k \plus{} 1} \minus{} b_{k \plus{} 1}}{ \minus{} \frac {a_{k \plus{} 1}M^{k \plus{} 1}_{a}}{a} \plus{} \frac {b_{k \plus{} 1}M^{k \plus{} 1}_{a}}{a}} \equal{} \frac { \minus{} a}{M^{k \plus{} 1}_{a}}$ easly seen that (as didilica show $ M(a) \equal{} \frac {1}{a}$ Hence limit is equal to $ \minus{} a^{k \plus{} 2}$)
we are done.
Now I have no idea How   without calculation can be shown this result , I will glad if fedja help  :)maybe my solution looks difficult but there is no nothing difficult it's easy to understand than to explain

I guess it's time to post the solution. First, the geometric one. Let's draw the picture:
[asy]size(200);
pointfontsize=8;

D((0,10)--(0,0)--(10,0), black);
path P=MC("f",D((2,1)..(5,3)..(8,6)),0.2,S), Q=MC("g",D((2,1.1)..(5,3.4)..(8,6.7),green),0.2),
     V=(5,0)--(5,10), H=(0,5)--(10,5);
pair X=D((5,0),black), Y=D((0,5),black),
     A=MP("A",D(IP(V,Q)),N), B=MP("B",D(IP(V,P))),
     C=MP("C",D(IP(H,Q))), D=MP("D",D(IP(H,P))),
     E=MP("E",D(IP(D--(D.x,10),Q)),N),
     Z=MP("y",D((D.x,0),black));
MP("x",X,S);
MP("x",Y,W);
DPA(X--A^^Y--D^^E--Z,black);[/asy]
The ratio we are interested in (up to minus sign) is $ |AB|/|CD|$. We would like to write it as $ \frac{|AB|}{|ED|}\cdot\frac{|ED|}{|CD|}$. The second ratio is just the slope of our curves near $ 0$, which is about $ a$ (the closer we are to $ 0$, the better is this approximation). To find the first ratio, just notice that it is about $ (x/y)^{d\plus{}1}$ (because $ (g\minus{}f)(x)\asymp cx^{d\plus{}1}$) and, clearly, $ x/y\approx a$ (slope considerations again). This yields the answer $ \minus{}a^{d\plus{}2}$ for the limit (in the formulation I posted; everybody else tried to compute the limit of the inverse fraction :roll: ). When $ a\equal{}1$, you get $ \minus{}1$ regardless of $ d$.

What Extremal tried to do also leads to a solution. Indeed, if $ f$ and $ g$ coincide up to order $ d$, then so do $ f^{\minus{}1}$ and $ g^{\minus{}1}$. The equations for their $ d\plus{}1$-st coefficients are
$ a_{d\plus{}1}/a^{d\plus{}1}\plus{}a\alpha_{d\plus{}1}\plus{}\text{junk}\equal{}0$
and
$ b_{d\plus{}1}/a^{d\plus{}1}\plus{}a\beta_{d\plus{}1}\plus{}\text{junk}\equal{}0$
where $ \text{junk}$ is quite complicated but [b]the same[/b] in both equations. This allows to conclude that $ \alpha_{d\plus{}1}\minus{}\beta_{d\plus{}1}\equal{}(b_{d\plus{}1}\minus{}a_{d\plus{}1})/a^{d\plus{}2}$,
which leads to the same answer.

As to "instantly", I'm inclined to agree with jmerry :).


Is this statement true?
"If $ a^3\equal{}9k$ and $ a$ is a 3 digit integer, $ a$ is the multiple of $ 9$"

Not true.

When $ k\equal{}111111\frac{1}{9}$ and $ a\equal{}100$

Presumably $ k$ is an integer.  But it's still false; $ a$ need only be divisible by $ 3$.


probability
ratio
trigonometry
function
geometric series
geometry
similar triangles

yo that was pretty hard today for some reason, i think i only got a 3... howd yall get the last three?

Please don't discuss the Math League questions for another day or two, because the contest is still being administered nationwide.

Well, I can't remember the fourth problem, but I can basically remember how I did the fifth and sixth ones (I got a perfect six this time   :) ).  So, for the fifth problem with that triangle, I easily found that that portion of the hypotenuse from the base up to its intersection with the altitude shown in the diagram had length 5/6 that of the total length of the hypotenuse.  I divided the base by the hypotenuse and then divided the base by 5/6 the hypotenuse so that, using similarity of triangles, I got cosx=5/(6cosx) where x is the smallest angle (on the right in the diagram) so that cosx=sqrt(5/6).  From this, sinx=sqrt(1-(cosx)^2)=sqrt(5), so the longer leg divided by the shorter leg=cosx/sinx=sqrt(5).

For the sixth problem, I took (1/2)(1/6)+(1/4)(5/6)(1/6)+(1/8)((5/6)^2)(1/6)+... or something like that, i.e. I found the probability that the coin didn't land heads and the dice rolled a two (or whatever number it was supposed to roll) on the first time, plus probability that coin didn't get heads twice and the dice didn't roll a two the first time but rolled a two the second time, plus that the coin doesn't get heads three times and . . ., then summing the geometric series I got 1/7 for the total probability.  There's probably a simpler solution but I didn't want to waste time looking for it when I knew how to do it this way.

I know that these explanations are really poorly written but I'm kinda in a hurry, sorry.

Number 5 was easy. The leg is the geometric mean of the hypothenuse and the closer segment of the hypothenuse the height cuts off. If the hypothenuse is length 1, the short leg is sqrt(1/6) and the long leg is sqrt(5/6). Therefore the ratio is sqrt5.
Basically, use similar triangles, no trig necessary. Using the geometric means means no Pythagorean theorem either.

That's a nice solution for number five, bictor.  It took me awhile to do it in terms of trig functions (not that I felt rushed during the contest) and I would've saved alot of trouble if I'd thought of your method.


[hide]81649
If you start with a lower two digit perfect square, you will be forced to end the number with fewer digits than 5:
649
49
364
1649
[/hide]

[hide="Solution"]
We know the first digit must be a 1, 2, 3, 4, 6, or 8.  The rest of the digits must be a 6, 5, 9, 4, or 1.  This means the middle digits must be an [b]intersection[/b] [size=59][color=red](thanks Farenhajt)[/color][/size]of these sets:

1st digit = 1, 2, 3, 4, 6, 8
Middle digit = 1, 4, 6
Last digit = 1, 4, 5, 6, 9

Now we look at it by choosing a first digit, which will imply a middle digit, which will imply another middle digit, until we can't use another middle digit, and are forced to use an end digit:

1st digit 1: 1649
1st digit 2: none
1st digit 3: 3649
1st digit 4: none
1st digit 6: 649
1st digit 8: 81649

Thus, the biggest possible number is $\boxed{81649}$

Yay, the first AIME problem that I could do immediately after looking at it![/hide]

[quote="mysmartmouth"]This means the middle digits must be a [b]union[/b] of these sets.[/quote]

Rather intersection.

The two digit squares are 16, 25, 36, 49, 64, and 81. So if 16 occurs in the number, then it must be followed by 4, 4 must be followed by 9 etc...
By this reasoning the longest number is

81649.

[quote="rnwang2"]Find the largest positive integer such that each pair of consecutive digits forms a perfect square (ex. 364).[/quote]

nvrmind edited.

[quote="t0rajir0u"]Uh, this can't be an AIME problem if the answer is five digits long.[/quote]

they probably asked for the remainder if the number is divided by 1000.

Um... I don't think they would do that in this case because look at the last 3 digits of every possibility in mysmartmouth's post. Although, they could ask for the sum of the digits or something else that could make it a valid answer...

I think this problem is way too easy to be in AIME... Maybe it was on AMC or something... Oh, wait... AMC's only have  multiple choice... MathCounts maybe?

the problem may have been reworded. but it was the first question on AIME2 2001.

[hide="A bit late but still good"]Since we don't have a 2 digit square greater than 81, then 81 would be our first two digits. We are looking for the greatest 2 digit square between 10 and 19 and that's 16. Again for 6, => 64 and again for 4 => 49.

The number is $81649$[/hide]


This is a problem I made  for my son long ago.  I think It is original in the sense that I have never seen it  in this form before anywhere , yet the theme is fairy common and since it is a nice looking problem, I am sure it must have been used by others  some place.  (I have posted it in mathcounts forum long ago)

My seven digit telephone number has an interesting property:    If you add the  square of the number formed by the first 3 digits to the  square of the  number formed by the last 4 digits, you  get my original phone number. What is the this phone number?

( To have fun  Computer  programs / calculators are not allowed)

Looks like this is  moved from  intermediate form .. any way here is hint:
10^8+1  is divisible by 17 (=4^2+1^2).

Let n be the 7-digit number, written as 10000 a + b
with 100 <= a <= 999 and 0 <= b <= 9999.

The stated condition amounts to
   n = 10000 a + b = a^2 + b^2
which is equivalent to
   (2a - 10000)^2 + (2b - 1)^2 = 10000^2 + 1 = N

This means we are looking for another decomposition of N as a sum of two squares (i.e. different from 100000^2+1^2)

Since N is a multiple of 17 and N is equal to 1 (mod 4), we have that N = 17 M with M = 1 (mod 4), which implies by a well-known theorem that M is also a sum of two squares : letting M  = x^2 + y^2, we have now
   N = (4^2 + 1) (x^2 + y^2)

We now use the well-known formula
(p^2+q^2) (r^2+s^2) = (pr + qs)^2 + (ps - qr)^2 = (pr - qs)^2 + (ps + qr)^2
with p=4 and q=1, and try to identify one the sums of squares with 10000^2+1
(4^2+1^2) (r^2+s^2) = (4r + s)^2 + (4s - r)^2 = (4r - s)^2 + (4s + r)^2
First case :
4r + s = 10000 and 4s - r = 1 => r = 39999/17, not an integer solution
Second case :
4r + s = 1 and 4s - r = 10000 => r = -588 and s = 2353 => the second decomposition becomes 10000^2 + 1 = (-4705)^2 + (8824)^2
(all the other possibilities, with minus signs or the second decomposition are equivalent up to a sign change)
We can now assume 2a - 10000 = -8824 and 2 b - 1 = 4705, to find a = 588 and b = 2353, and thus n = 5882353.

Some additional remarks :
 - all of this was done without having to factor M (which is actually a prime)
 - what I still do not understand is why the a and b we find in the end are precisely equal to r and s (up to a sign change) or, equivalently, why M is equal to a^2+b^2.  I would be glad if anybody could explain this to me ...

Sorry in a hurry but would post if need be..

Nice solution.

Yes, it is rather fun to see the number is M, it is not obvious (well it wasn't to me at first  when I made the problem, and was surprised to see this too) but can be shown that it will always be  M*k^2 where k  is one of the two squares which appears in  the first factor is  (here in 17 as 4^2+1^2).  So other such number would be ()Though > 10^7)   17*2353^2 =  94 12,2 353 = 9412^2+2353^2.

I really liked this problem and got an excuse to teach some fun stuff to my son..


Hi all,

I enjoy solving math problems as well as chess/go problems, but (inevitably) come across problems where I have no idea how to proceed. Usually I let the problem simmer for a while, and if I can't think of a solution after 2-3 days I ask my teacher/consult the solution. However most of the time I feel disappointed when I realise that the solution was within my grasp and it took just a bit more work to be able to see it.

How do you guys define 'hard work'? For you, when does something challenging become a sheer waste of time? How does one maintain a good balance between banging one's head against the wall and sanity =) ?

Would like to hear your thoughts
Thanks!

Everybody has a different patience level.  For me, I can tackle a problem many times, 30 minutes to an hour each time, and not make any progress for days, but eventually I will find the solution.

I do not have the patience to just work for 2+ hours on one problem, I would get frustrated and my brain would stop working.

Working on it in small intervals works for me, but sometimes I still can't get the answer  :blush:

I tend to solve problems by thinking up a strategy, running through it, seeing if it works, and if it doesn't, coming up with another strategy.  The strategy can be something I've planned out well in advance, or it can just be, "I wonder what happens if I..."

I tend to "get stuck" on problems when my most recent strategy falls through.  Then I put aside the problem for twenty minutes, an hour, an afternoon, a few days, a week, or whatever it takes, while I put it on "simmer" on the back burner of my mind.  There is one problem I've been sitting on since December, and I think I'll finally be able to solve it once I have a week or so of free time to work on it.

Of course, on other problems, I just get frustrated and throw in the towel.  Some problems I have the patience to pursue for months or years.  Others, I give up after 30 minutes.

I find the failed strategies one of the more valuable parts of the problem-solving experience.  I analyze why the strategy didn't work here, and which conditions it would work under, and file it away in my mind for future use.  Sometimes I include this in my papers, because I think it's useful information.  Other people do, too.  I was once contacted by a group who thought that one of the strategies I discussed that didn't work for my data would be the only way to handle their data, and I agree with their assessment.

Something challenging only become a sheer waste of time when I don't find the problem interesting or enjoyable any more, or I'm learning nothing from the experience.  There are plenty of good problems to work on in the world.  Choose the ones you truly want to work on!

Edited to add:  Also, I hate working under time pressure.  If the time I have to complete a problem is relatively open-ended relative to the length of the problem, I'm fine.  When I'm under time pressure, and three or four approaches in a row don't work, I start to panic.  Then the panic makes things worse.  Going down blind alleys is part of the way I naturally solve problems; it's just demoralizing to work under the prospect that if the next approach doesn't work, bad things will happen to me.

WELL...

You should always, if you get stuck, go onto another math problem and come back to it...

That way you can always have a fresh perspective...

I usually can tolerate up to a week.. but that's it.. then i look in the back of the book/solutions manual/teacher.

Sometimes, you really do just need a break. I know everyone else has said this--but bear with me. If you can't think of a solution--let it sit. Sometimes, letting it "simmer" will help you (the whole subconscious thing), and [i]I[/i] know that my brain can get cramped. Another thing to definetely do is to just breathe.
This works for several reasons: (and there are more)
A) You'd die if you didn't,
B) Oxygen is great brain food, and
C) Hyperventilating just makes the panicking situation far wose than it seems or actually is.

Work on it later, when you feel like it. Sometimes, after you finish "letting it simmer", review in your mind. Remember what you did, and the new plans. Then sit down to the problem again and proceed to solving it. Solving a problem by yourself is far more satisfying than after you're asked for help. However, sometimes help is necessary, so you'll get it the next time. Ask if you need it, but really work for it fist. :)

hmmm... this has been interesting because i can do AMC12 and some AIME probs but usually within 5-10 mins. I don't think i've ever worked longer which i should be because i get frustrated with the problem and look at the solution and then i have the same prob that you had. I need more patience, i naiively assumed that most math geniuses could do any prob within 15 mins, i had no idea people actually took more than a day, i feel stupid. Anyways great posts this has opened my eyes a bit.

Thanks for the replies! If anyone has any more advice to offer please don't hesitate.

The advice has been encouraging. I think sometimes one shouldn't be too hard on oneself for asking for advice. Sometimes you learn something completely new, at other times you learn something psychological about yourself i.e. impatience, or that you are weak at a certain problem solving technique.

And yes, if you think about it, even great geniuses take days/months to solve problems that now appear trivial. After all, it's because they solved them that the solutions now seem obvious. :)

When I was in high school I used to keep a notebook of problems that I couldn't solve, but knew were within "high school" ability. I started it as a freshman in high school, and over time I would add problems to it and solve problems from it. Sometimes I would have new insights (or have learned something new) that allowed me to solve problems that I hadn't been able to solve before.

By not looking up the solutions (which I often did not have anyway), it helped me to become more creative in my problem solving because I had to try many ways sometimes before I found the solution. The downside was, even when I did solve a problem it may not have been by the most effective method. Sometimes it might have been just brute force where an official solution might have shown me a more clear way. But it basically ended up being a collection of some of my favorite problems from high school. I wish I still knew where it was. :(


modular arithmetic
partition
number theory
Product
IMO
IMO 1970

Find all positive integers $n$ such that the set $\{n,n+1,n+2,n+3,n+4,n+5\}$ can be partitioned into two subsets so that the product of the numbers in each subset is equal.

We claim that no such $n$ exists.

Before we present the proof, we define some notations for the sake of convenience and clarity. Let $A$ denote the set $\{n, n+1, \ldots, n+5 \}$. We denote $f(X)$ as the product of all the elements in $X$, defined only when $X$ is a set whose elements are integers.

Solution 1 (a more combinatorial approach):

Suppose such an $n$ that can be partitioned exists. Let $A$ be partitioned into subsets $B$ and $C$ such that $f(B) = f(C)$.

Note that for a partition to be possible, we must have for each $i \in \{0, 1, \ldots, 5\}$ such that $n+i \equiv 0 \pmod{p}$ for some prime $p$, $\exists j \neq i, j \in \{0, 1, \ldots, 5\}$ satisfying $n+j \equiv 0 \pmod{p}$, since otherwise exactly one of $f(B), f(C)$ is $\equiv 0 \pmod{p}$, which is a contradiction as our assumption that $f(A) = f(B)$ implies $f(A) \equiv f(B) \pmod{p} \forall p$. This immediately restricts the possible prime divisors of the integers in $A$ to $2, 3$, and $5$, since if $\exists$ some prime divisor $p > 5$ dividing $n+i$, we have $n+i+p > n+5$ and $n+i-p<n$, implying none of the remaining integers in $A$ would be $\equiv 0 \pmod{p}$.

Since $f(A)$ is a product of six consecutive integers, at most $\lceil \frac{6}{5} \rceil = 2$ of the integers in $A$ are $\equiv 0 \pmod{5}$, and at most $\lceil \frac{6}{3} \rceil =2$ of the integers in $A$ are $\equiv 0 \pmod{3}$. From the above argument, it follows that $\geq 2$ of the integers in $A$ are divisible by $5$, and that $\geq 2$ of the integers in $A$ are divisible by $3$, implying that $A$ has exactly two multiples of $5$ and two multiples of $3$. We note that the two multiples of $5$ are of different parities, and similarly, the two multiples of $3$ are also of different parities. We note also that there are exactly $3$ odd integers in $A$, thus one of them is a multiple of $3$, one of them is a multiple of $5$, implying at least one of them cannot have any odd prime divisors, which forces it to be $1$. This implies $n=1$, but $A = \{1, 2, \ldots, 6 \}$ doesn't work (Contradiction).

Solution 2 (a more algebraic approach):

Suppose such an $n$ that can be partitioned exists. Let $A$ be partitioned into subsets $B$ and $C$ such that $f(B) = f(C)$. We have $f(A) = {[f(B)]}^2$, a perfect square. We note that there must be $\geq 2$ multiples of $5$, since otherwise we would have exactly one of $f(B), f(C) \equiv 0 \pmod{5}$, which is a contradiction as $f(B) = f(C)$ implies $f(B) \equiv f(C) \pmod{5}$. It follows then that $n \equiv n+5 \equiv 0 \pmod{5}$. We note that $((n+2)^3)^2 < f(A) < ((n+3)^3)^2$, thus $(n+2)^3 < f(B) = f(C) < (n+3)^3$. WLOG, assume $n+2 \in B$.

Suppose $n+3 \in B$.
We have $n+1 < \frac{(n+2)^2}{n+3} < \frac{f(B)}{(n+2)(n+3)} < \frac{(n+3)^2}{n+2} < n+5$, which implies $\frac{f(B)}{(n+2)(n+3)} = n+4$. Yet, $f(C) = n(n+1)(n+6) < (n+2)(n+3)(n+4) = f(B)$. (Contradiction) Thus $n+3 \in C$.

Note that $\frac{f(B)}{n+2} > (n+2)^2$ and that $(n+2)^2 > n(n+4) > n(n+1)$. Note also that both $n$ and $n+5$ cannot be both in the same subset. Thus, $n \not\in B \Rightarrow n \in C, n+5 \in B$. Obviously $(n+2)(n+4)(n+5) > n(n+1)(n+3)$, thus we are left with the only possibility of $(n+1)(n+2)(n+5) = f(B) = f(C) = n(n+3)(n+4)$, which reduces to the equation $n^2 + 7n + 10 = 0$. This has obviously no positive integer solutions. (Contradiction)

If any number in the set has a prime in its prime factorisation greater than 5, that prime cannot appear in the factorisation of another number, so we come to a contradiction. So all numbers are of the form $2^n.3^m.5^k$.

If there were 6 such consecutive numbers, then modulo 5 our set is $0 1 2 3 4 0$. Now of those 4 non-zero numbers, at most 2 are 0 mod 2, and the other 2 cannot be both 0 mod 3, so at least one of them is not divisible by 2,3 or 5, but is greater than 1 thus has a prime factor greater than 5, contradiction.

This is exactly Solution 1 in my earlier post.

Ah, I'm sorry I didn't see it were 2 solutions. :oops: I thought it was just such a long solution so I didn't read it

This problem could be solved in much nicer and more number theory way....

I suppose that those subsets [b]don't have common elements[/b], otherwise it is stupid for solving :).

So let $M=\{n, n+1,n+2,n+3,n+4,n+5\}$ and $A$ and $B$ are subsets of $M$ s.t. $A\cap B=\phi$ and $A\cup B=M$ and $P(A)=P(B)=y$ where $P(X)$ is product of elements in subset $X$.
Prime number $7$ can divide only one or none number from this set, so if $7$ divides some number then $7|P(A)$ and $not(7|P(B))$ so $P(A)$ and $P(B)$ cant be equal, from this we see that $7$ don't divide any number from $M$ which implies that $7|n+6$ and $7|n-1$.
Notice that $P(M)=y^2$ and $P(M)\equiv_7 1\cdot 2...\cdot 6\equiv_7 -1$ but square can't be congruent with $-1$ by module $7$ so contradiction!! :)

That's a brillant solution [i]M@re[/i]!  :lol:
Well, so we have three different approaches (combinatorial, algebraic, number theoretic) to this problem. Problems like these are beautiful as they can be solved in various ways, but they also make the categorization of problems into combinatorics, algebra, or number theory forums a hard choice. haha..  :P

Sorry to revive, but I have another idea.

[hide="Main idea"]$ (n\plus{}3)|n(n\plus{}1)(n\plus{}2)(n\plus{}4)(n\plus{}5)(n\plus{}6)$ and $ (n\plus{}4)|n(n\plus{}1)(n\plus{}2)(n\plus{}3)(n\plus{}5)(n\plus{}6)$[/hide]
[hide="Implication"]$ (n\plus{}3)|36$ and $ (n\plus{}4)|48$[/hide]
[hide="Conclusion"]No solution[/hide]

Please correct me if I'm wrong.

Mod 5 implies that $ n$ and $ n\plus{}5$ are in separate subsets.
Suppose the subset with $ n$ had four elements, then we require $ n(n\plus{}1)(n\plus{}2)(n\plus{}3) \leq (n\plus{}5)(n\plus{}4)$ which can only occur for $ n\equal{}1$, contradiction. If each set has three elements, we require

$ n(n\plus{}4)(n\plus{}3) \geq (n\plus{}5)(n\plus{}1)(n\plus{}2)$ or $ n(n\plus{}4\plus{}1)(n\plus{}3) > (n\plus{}5)(n\plus{}1)(n\plus{}2)$ or $ n^2\plus{}3n > n^2\plus{}3n\plus{}2$ contradiction.

[quote="M@re"]I suppose that those subsets [b]don't have common elements[/b], otherwise it is stupid for solving :).

So let $M=\{n, n+1,n+2,n+3,n+4,n+5\}$ and $A$ and $B$ are subsets of $M$ s.t. $A\cap B=\phi$ and $A\cup B=M$ and $P(A)=P(B)=y$ where $P(X)$ is product of elements in subset $X$.
Prime number $7$ can divide only one or none number from this set, so if $7$ divides some number then $7|P(A)$ and $not(7|P(B))$ so $P(A)$ and $P(B)$ cant be equal, from this we see that $7$ don't divide any number from $M$ which implies that $7|n+6$ and $7|n-1$.
Notice that $P(M)=y^2$ and $P(M)\equiv_7 1\cdot 2...\cdot 6\equiv_7 -1$ but square can't be congruent with $-1$ by module $7$ so contradiction!! :)[/quote]
This is a valid solution for $7$ replaced by any prime $p\equiv 3\mod 4$.

By the way, where's the combinatorics in this problem?  :huh:
[color=#008000]Moderator says: indeed.[/color]

simple solution.If we prove,that one of these numbers,have a prime divisor $ P \ge 7$ than it will be the only one,which is divisible by $ P $.We have exactly $ 3 $ even and $ 3 $ odd numbers,in these $ 3 $ odd number,at most $ 1 $ of them is divisible by $ 3 $ and at most $ 1 $ is divisible by $5$,which means the left odd number is divisible by a prime number,distinct from $ 2,3,5 $,which means it's divisible by $ P \ge 7 $ . q.e.d. so there is no solution

[quote="HelloWorld"]How do we prove the generalized of this problem ?
I have seen it from IMO Compendium Page 372 on its Remark. :
" Remark. Erdo˝s has proved that a set $ n, n \plus{} 1, . . . , n \plus{} m$ of consecutive natural numbers can never be partitioned into two subsets with equal products of elements.  :)[/quote]
i think it will have the same solution,that is,there exists such $ P $ prime,which divides only one element of the set

I know contestant in country anywhere solved generalization version.Greece？

[hide=My solution]There are $6$ consecutive numbers, so there must be at least one number which is divisible by $5$. If such partition exist, there must be two number divisible by $5$. And hence, they must be $n$ and $n+5$. Since the condition, we have to put these to numbers into two different subsets. If the sets have different caridnality, there will be one set with product greater than the other. Hence, both of the sets must have cardinality of $3$. Thus, we have the following cases:

1) $A=\{n,n+1,n+3\}$ and $B=\{n+2,n+4,n+5\}$. This obviously is not the desired partition, as $n<n+1, n+2<n+3$ and $n+4<n+5$.

2) $A=\{n,n+1,n+2\}$ and $B=\{n+3,n+4,n+5\}$. Fails for the same reason as the first case.

3) $A=\{n,n+2,n+3\}$ and $B=\{n+1,n+4,n+5\}$. Fails for the same reason as the first case.

4) $A=\{n,n+2,n+4\}$ and $B=\{n+1,n+3,n+5\}$. This also fails, since \[ n(n+2)(n+4)=(n+1)(n+3)(n+5) \]can be simplified to $n^2+5n+10=0$, which has no zero.

And hence, there is no such $n$.
 [/hide]

[quote=div5252]At least one of six consecutive numbers is divisible by 5. From
the given condition it follows that two numbers must be divisible by 5.
These two numbers are necessarily n and n + 5. Therefore n and n + 5
are in distinct subsets. Since n(n + 1) > n + 5, it follows that a required
partition cannot be considered with subsets of different cardinality. Thus
each subset must contain three numbers. The following possibilities have
to be considered:
a) {n, n + 2, n + 4} ∪ {n + 1, n + 3, n + 5}
b) {n, n + 3, n + 4} ∪ {n + 1, n + 2, n + 5}.
In case a), n < n + 1, n + 2 < n + 3 and n + 4 < n + 5.
In case b), the condition of the problem gives:
n(n + 3)(n + 4) = (n + 1)(n + 3)(n + 5).
We obtain n2 + 5n + 10 = 0 and this equation has no real solution.[/quote]

Direct landed from Titu's book

[b] Lemma : [/b] Let $ p = 4k+3 $ be a prime. Then the set $\mathcal{A} =\{ n,n+1,\dots,n+p-2\} $ cannot be split into two disjoint subsets such that their product are the same.

[b] Proof : [/b]
Suppose on the contrary it is possible to partition $\mathcal{A}$ into two disjoint subsets, $ \mathcal{B}, \mathcal{C} $ such that $ \prod_{x \in \mathcal{B} } x = \prod_{ x \in \mathcal{C} } x = d $
Observe that $\not \exists \ i \in \{0,1,\dots,p-2\} $ such that $ p \ | \ n+i $
If $ p \ | \ n+i $ for some $ i \in \{ 0,1,\dots,p-2\} $
Then by problem condition there must $ \exists \ j \in \{0,1,\dots,p-2\}, j \not= i $ such that $ p \ | \ n+j $
WLOG , let $ i < j $, then $ n+i < n+j < n+i+p $
$ \Rightarrow pm < n+j < p(m+1) $ where $ pm = n+i \Rightarrow p \not | \ n+j $
Hence $ \not \exists \ i \in \{0,1,\dots,p-2\} $ such that $ p \ | \ n+i $

Hence the set $\mathcal{A}$ forms a permutation of $\{1,2,\dots,p-1\} \ ( \mbox{ mod } p ) $
Then $ \left( \prod_{x \in \mathcal{B} } x \right) \left( \prod_{ x \in \mathcal{C} } x \right) = d^2 $
$$ \Rightarrow \prod_{x \in \mathcal{A} } x = d^2 $$
$$ \Rightarrow \prod_{i=1}^{p-1} i \equiv d^2 ( \mbox{ mod } p ) \Rightarrow (p-1)! \equiv d^2 ( \mbox{ mod } p ) $$
Hence by Wilson's Theorem $ -1 \equiv d^2 ( \mbox{ mod } p ) $
Thus $-1$ is a QR $ \mbox{ mod } p $ hence $ \left( \frac{-1}{p} \right) \equiv 1 ( \mbox{ mod } p ) $
But $ \left( \frac{-1}{p} \right) \equiv ( -1 )^{\frac{p-1}{2}} \equiv -1 ( \mbox{ mod } p ) $( Contradiction )
Hence $ \not \exists \mathcal{B}, \mathcal{C} \subset \mathcal{A} $ such that $ \mathcal{B} \cap \mathcal{C} = \phi $, $ \mathcal{B} \cup \mathcal{C} = \mathcal{A} $ and $ \prod_{x \in \mathcal{B} } x = \prod_{ x \in \mathcal{C} } x $
Hence our lemma is true.

This problem is the particular case of $ p=7 $.

my solution needs a little hard work:
consider the set ${n,n+1,n+2,n+3,n+4,n+5}$
so we now that at least one of them is divisible by $5$ so if there is just one number divisible by $5$ we can't partition the set so there is exactly two numbers divisible by $5$ and obviously those numbers are $n$ and $n+5$ so we put each one in a different partition so now we should solve a bunch of quadratic equations

$n\left(n+4\right)\left(n+3\right)=\left(n+5\right)\left(n+1\right)\left(n+2\right)$ ,answers: $n=-\frac{5}{2}+i\frac{\sqrt{15}}{2},\:n=-\frac{5}{2}-i\frac{\sqrt{15}}{2}$
$n\left(n+1\right)\left(n+3\right)=\left(n+5\right)\left(n+4\right)\left(n+2\right)$,answers: $n=\frac{-35+\sqrt{105}}{14},\:n=\frac{-35-\sqrt{105}}{14}$
$n\left(n+1\right)\left(n+4\right)=\left(n+5\right)\left(n+3\right)\left(n+2\right)$,answers: $n=\frac{-27+\sqrt{129}}{10},\:n=\frac{-27-\sqrt{129}}{10}$
$n\left(n+1\right)\left(n+4\right)=\left(n+5\right)\left(n+3\right)\left(n+2\right)$,answers: $n=\frac{-27+\sqrt{129}}{10},\:n=\frac{-27-\sqrt{129}}{10}$
$n\left(n+3\right)\left(n+2\right)=\left(n+5\right)\left(n+1\right)\left(n+4\right)$,answers: $n=\frac{-23+\sqrt{129}}{10},\:n=\frac{-23-\sqrt{129}}{10}$
$n\left(n+4\right)\left(n+2\right)=\left(n+5\right)\left(n+1\right)\left(n+3\right)$,answers:$n=\frac{-23+\sqrt{129}}{10},\:n=\frac{-23-\sqrt{129}}{10}$

so non of them gives an integer solution also it is obvious that $n\left(n+1\right)\left(n+2\right) \leq \left(n+5\right)\left(n+4\right)\left(n+3\right)$
so we don't check the situation $n\left(n+1\right)\left(n+2\right)=\left(n+5\right)\left(n+4\right)\left(n+3\right)$

so it has no solution :oops:

No $n.$

Clearly $n\equiv 1\pmod{7}$ otherwise one of these terms is divisible by $7$ (and none of the rest are), so one subset is divisible by $7$ and the other isn't. Now note $6!\equiv -1\pmod{7},$ and $A^2\equiv -1\pmod{7}$ has no solutions by exhaustion.

Here is a fairly dumb solution.
The answer is that there does not exist such an $n$.
Consider an arbitrary fixed $n$. Then, taking the entire set $\pmod{n+2}$ we see that we have $\{-2,-1,0,1,2,3\}$. Thus one of the subsets from the partition must be congruent to $0 \pmod{n+2}$. The other subset must be congruent to a number dividing $(-2)(-1)(1)(2)(3)=12 \pmod{n+2}$. However since these subsets should have equal product, it follows that $12 \equiv 0 \pmod{n+2}$. This, along with the fact that $n$ is a positive integer, implies $n \in \{4,6,12\}$.
For $n=4$ the set becomes $\{4,5,6,7,8,9\}$. However we see that exactly one of the sets is divisible by $5$ and therefore $n=4$ cannot be partitioned.
For $n=6$ the set becomes $\{6,7,8,9,10,11\}$. However we see that exactly one of the sets is divisible by $13$ and therefore $n=6$ cannot be partitioned.
For $n=12$ the set becomes $\{12,13,14,15,16,17\}$. However we see that exactly one of the sets is divisible by $17$ and therefore $n=12$ cannot be partitioned.
Putting these together, it follows that no $n$ exists. $\blacksquare$

Very easy problem.

[hide=Storage]
I claim that no such $n$ exist. For convenience, let $f(n)=n(n+1)(n+2)(n+3)(n+4)(n+5)$. Suppose, for the sake of contradiction, that such a positive integer $n$ does exist. Then, we have $P(S_1)=P(S_2)$ where $P(S)$ denotes the product of all elements in a set $S$ and $\{S_1, S_2 \}$ is a partition of $\{n, n+1, n+2, n+3, n+4, n+5 \}$. Note that either $n \in S_1$ or $n \in S_2$ which gives that either $n \mid P(S_1)$ or $n \mid P(S_2)$ but since $P(S_1)=P(S_2)$, we have that $n^2 \mid P(S_1)P(S_2)=n(n+1)(n+2)(n+3)(n+4)(n+5)$. Thus, $n \mid f(n) \Longrightarrow n \mid 120$ which gives $n \in \{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 \}$. Note that $f(n)$ must be a perfect square, which we can check is not achieved since in each of the cases, atleast one odd prime $p \geq 5$ has $\nu_p(f(n))=1$ which is not possible since $f(n)$ is a perfect square. A contradiction. Thus, no such $n$. $\blacksquare$
[/hide]

The idea is to take mod $7$; notice that if one of $n, n+1, n+2, \cdots, n+5$ is a multiple of $7$, it is the sole multiple of $7$, so the partition is impossible. Else, if $n \equiv 1 \pmod 7$, then notice that the two products $x \equiv y \pmod 7$ of the two sets. But this implies that $xy \equiv x^2 \equiv -1 \pmod 7$, which is impossible.

[color=#9a00ff][b]Remark.[/b][/color] Actually, one can show the stronger problem that there can never exist two subsets of the set that have equal product (dropping the partition condition).

[url]https://users.renyi.hu/~p_erdos/1975-46.pdf[/url]

Then we're done.

Solution with $5$ replaced by $1997$, and the set denoted by $A$.

$\boxed{\text{No such }n\text{ exist.}}$ The following proof hinges on the fact that $1997$ and $1999$ are prime. Clearly, there is a multiple of $1997$ in $A$, and said multiple is unique unless $1997$ divides both the extrema $n,n+1997$. In any case, there cannot be greater than two subsets in a partition of $A$, or else some but not all of the subsets contain an element divisible by $1997$, preventing the subset products from being all equal (mod $1997$). So if FTSOC $A$ is partitionable in the given manner, it can only be partitioned into two subsets, making the product of $A$'s elements a perfect square $a^2$.

Now observe that $A$ contains a unique multiple of $1999$, unless $A\equiv Z_{1999}^\times\pmod{1999}$. But if $A$ contains a unique multiple of $1999$, only one of its two partition subsets can have product divisible by $1999$, contradiction; so we indeed have $A\equiv Z_{1999}^\times$. Then the product of $A$'s elements is $a^2\equiv1998!\equiv-1\pmod{1999}$, which is a contradiction because $(\tfrac{-1}{1999})=(-1)^{(1999-1)/2}=-1$. So the initial assumption was faulty, and $A$ cannot be partitioned in the specified manner, as claimed. $\blacksquare$

Note that $n \equiv 1 \pmod 7$ because if $n$ is any other residue modulo $7,$ then one of the number is a multiple of $7,$ which clearly create different products for any partition. From Wilson’s, $6! \equiv -1 \pmod 7.$ Let one of the partitioned subset be $a$ and other be $b.$ Then $$a \equiv b \pmod 7 \implies ab \equiv a^2 \equiv 6! \equiv -1 \pmod 7.$$ However, $-1$ is not a quadratic residue modulo $7,$ and hence we cannot partition the set into two subsets so that the product of the numbers in each subset is equal. $\blacksquare$

The answer no such $n$. Assume the subsets are nonempty, because otherwise the products cannot be equal.

If at least one of these numbers are divisible by $7$, then exactly one of them is divisible by $7$. This implies that exactly one subset has a product of a multiple of $7$, which means the product of numbers in each subset cannot be equal.

Otherwise, the set is equivalent to $\{1,2,3,4,5,6\}$ modulo $7$. Notice that the product of the numbers in the set must be a perfect square. So the product of numbers in the set is $6! = 720\equiv -1\pmod 7$. Since $7\equiv 3\pmod 4$, $-1$ is not a quadratic residue modulo $7$, so the product cannot be a perfect square, absurd.

I claim that there is no such $n$.
Define $\Pi X$ to be the product of all the elements in a set $X$. Observe that there must always be exactly three even integers. We have two separate cases when these respective even integers are $\{n, n+2, n+4\}$ and $\{n+1, n+3, n+5\}$.

[b]Case 1:[/b] $\left(\{n, n+2, n+4\} \right)$. Then assume $n = 2^km$ for some odd $m$ and integer $k$. We then have $n + 2 = 2(2^{k - 1}m + 1)$ and $n + 4 = 2^2(2^{k-2} + 1)$. Therefore we must have $n+2, n+4$ to be in a different set from $n$. Without loss of generality, assume $n \in A, \{n+2, n+4\} \in B$. Therefore $k = 3$.
Now observe that within six consecutive integers there must be a multiple of $5$. Hence as $\Pi A = \Pi B$, we must have $n$ and $n+5$ as these multiples of $5$. As already $n \in A$, we must have $n + 5 \in B$. We thus have Also as $\deg (\Pi A) = \deg (\Pi B)$ (else we trivially obtain strict inequality), we must have the following: \[n(n+1)(n+3) = (n+2)(n+4)(n+5).\] However $n < n+2, n+1 < n+4$ and $n + 3 < n + 5$. Hence no solution exists.

[b]Case 2:[/b] $\left(\{n+1, n+3, n+5\} \right)$. If we assume once more that $n + 1 = 2^km$ for an odd integer $m$, and $k$ is an integer. We must similarly have $k=3$ and $\{n, n+5\}$ as multiples of $5$. Hence if we assume $n+1 \in A$ and $\{n+3, n+5\} \in B$, we must have the following cases:
\begin{align*}
&(n+1)n(n+2) = (n+3)(n+5)(n+4) \\
&\text{or} \\
&(n+1)n(n+4) = (n+3)(n+5)(n+2) \\
&\text{or} \\
&(n+1)(n+2)(n+4) = (n+3)(n+5)n
\end{align*} The first two equalities are trivial simply by observing that the RHS is always strictly larger than the LHS by comparing factors. The final equation upon simplification yields $n^2 + n = 8$, which provides no integer solutions. Hence no integers solutions exist to the initial problem posed. $\blacksquare$

We claim that there exists no such $n$.

 Let the two sets be $S_1$ and $S_2$. Let the product of the elements of each set be $P_1$ and $P_2$ respectively.\\

    It is clear that $n=1$ doesn't work since $1+4=5$ is in at most one of $S_1$ and $S_2$ which makes one one $P_1$ and $P_2$ divisible by 5 and not the other, making it impossible that they are equal. Assume WLOG, $|S_1|\geq |S_2|$. Then, if $|S_1|\geq 4$ we have,
    \[P_1 \geq n(n+1)(n+2)(n+3)>(n+4)(n+5)\geq P_2\]
    So, $|S_1|=3$. This means we have three elements each in both the sets. Now, if note that if $n\not \equiv 0 \pmod{5}$, only one of $n,n+1,n+2,\dots, n+5$ is divisible by 5, making it impossible for the two products to be equal. Thus, $n\equiv 0 \pmod{5}$ and $n$ and $n+5$ are in two distinct sets. Assume WLOG, $n \in S_1$. Now, say $S_1={n,a_2,a_3}$ and $S_2={b_1,b_2,n+5}$. If $a_2 \leq n+2$ we have that,
    \[P_1\leq n(n+2)(n+4) < (n+5)(n+3)(n+1) \leq P_2\]
    for all $n \geq 2$. So, $a_2 \geq n+3$ which means we have $a_2=n+3$ and $a_3=n+4$. This means,
    \[n(n+3)(n+4)=P_1=P_2=(n+1)(n+2)(n+5)\]
    which one can rearrange to obtain $n^2+5n+10=0$ which clearly has no positive integer solutions.

 Thus, there exists no positive integers $n$ such that the set $\{n,n+1,n+2,n+3,n+4,n+5\}$ can be split into two disjoint subsets such that the products in these subsets are the same.


[quote="236factorial"]How many space diagonals does a cube have?[/quote]

[hide]4[/hide]


integration
function
logarithms
calculus
calculus computations

This is baby stuff compared to Misan problems. Here are two methods:

1) $ \sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n}{2n\plus{}1} \equal{} \int_0^1\frac{1}{x^2\plus{}1}\,dx$

2) Consider the sinus Fourier series of the function $ f(x) \equal{} \frac{\pi}{4}$

[quote="Carcul"]
1) $ \sum_{n \equal{} 0}^{\infty}\frac {( \minus{} 1)^n}{2n \plus{} 1} \equal{} \int_0^1\frac {1}{x^2 \plus{} 1}\,dx$
[/quote]

how you get this?

As the Captain Obvious said:
[hide]
\[ \int_0^1\frac{1}{x^2\plus{}1}\,dx\equal{}\int_0^1\sum_{n\equal{}0}^{\infty}(\minus{}1)^n x^{2n}\,dx\equal{}\sum_{n\equal{}0}^{\infty}(\minus{}1)^n \int_0^1 x^{2n}\,dx\equal{}\sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n}{2n\plus{}1}\]
[/hide]

In general, $ \sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n}{\alpha n \plus{} \beta} \equal{} \int_0^1\frac{x^{\beta \minus{}1}}{1 \plus{} x^{\alpha}}\,dx$. To prove this, start with an expression for $ \sum_{k\equal{}0}^{n\minus{}1}(\minus{}1)^kx^{k \alpha}$.

Define:

$ F(x) \equal{} \sum^\infty_{i\equal{}0}\frac{x^{2i}}{2i\plus{}1}\equal{}$

$ \equal{} \frac{1}{x}\sum^\infty_{i\equal{}0}\frac{x^{2i\plus{}1}}{2i\plus{}1}\equal{}$

$ \equal{} \frac{1}{x}\int\left(\sum^\infty_{i\equal{}0}x^{2i}\right)\,dx\equal{}$

$ \equal{} \frac{1}{x}\int\frac{dx}{1\minus{}x^2}\equal{}$

$ \equal{} \frac{1}{2x}\left(\ln(1\plus{}x)\minus{}\ln(x\minus{}1)\right)$

So:

$ F(\dot\imath)\equal{}\frac{1}{2\dot\imath}\left(\ln(1\plus{}\dot\imath)\minus{}\ln(\dot\imath\minus{}1)\right)\equal{}\frac{1}{2\dot\imath}\ln(\dot\imath)\equal{}\frac{\pi}{4}$


group theory
abstract algebra
superior algebra
superior algebra solved

Let M,N be normal subgroups of G such that G=MN. prove G/M x G/N isomorphic to G/(M^N) where ^ means intersection

Prove that the kernel of the canonical surjective map $G\to G/M\times G/N$ is $M\cap N$.

well, i thought i had it but then i'm having problems proving that the map which sends g to (gM,gN) is surjective.i want to know which element in G maps to the general element             (g_1 M,g_2N)? i tried this. since G=MN, there exists m,n s.t mn=1 so i fiddled around with combinations but i havent got it. i've been trying this for the last half an hr, someone give me a hint


[hide="hint"]The blocks one block in on each side will have no wax.[/hide]

[hide="answer"]The group of unwaxed blocks is contained in a large block 1 in by 2 in. by 4 in. Thus, its volume is 8 in.^3. Thus, there are 64 unwaxed blocks.[/hide]

[hide]With half-inch edges, you have side lengths of 10, 6, and 4. The ones that don't have any wax are the ones completely unexposed to the outside, meaning they have one cube above and below, to the left and right, and front and back. Subtract two from each of the side lengths and multiply: 8*4*2 = 64[/hide]

basically, all the outer blocks do not count, after taking those out, you have a smaller box, with sides two less than the previous (taking away a layer from the top and bottom), giving (10-2)(4-2)(6-2)=64

[quote="mr. math"]A block of cheese 5 inches by 2 inches by 3 inches is coated in a very thin layer of wax. The cheese is cut into 240 cubes with half-inch edges. How many cubes will not have any wax on them?

how would you do this problem?[/quote]

The only possibility for no wax is if they are in the interior of the box, so the answer is just $(10-2)*(6-2)*(4-2)=64$

[hide="A solution, not THE solution"]

The block, after it's cut into .5 inch cubes, has dimensions 10*4*6 half inches. We subtract 2 from each dimension so account for the wax:

$8*4*2=2^{3+2+1}=2^{6}=\boxed{\underline{64}}$[/hide]

[hide]
it is 10x4x6 cubes. the amt of cubes that dont have wax are 8x2x4, which is 64 cubes

[/hide]


here is the structure of an alkaloid prepared byby Richard Willstätter in 1901. It started from the seemingly related cycloheptanone, but required many steps and had an overall yield of only 0.75%.
in 1917, the same compound was prepared by robinson in just two steps from succindialdehyde, and a yield of 17% which with improvements was increased to 90%.
synthesise the the compound in just two steps using butanedial. you may use common orrganic reagents.

Double Mannich reaction. Ofcourse the original one required acetone but the latter gives more yield.

Yes, this synthesis is a classic and well known. Also, there is only one step involved, not two.

yes :)  almost seems like the whole wikipedia page has been posted here. neways to those interested, heres the link http://en.wikipedia.org/wiki/Tropinone


The question is:

How much genetic material is present in a cell during prophase 1 as compared to a cell that has completed meiosis II?

a.  one-quarter as much
b.  one-half as much
c.  twice as much
d.  the same amount
e.  four times as much

If I am reading the material right meiosis II has four daughter cells.  What I am not sure about is if there are even cells present during prophase I or if they come shortly after.

[hide]during prophase I, there is only one cell containing chromosomes in tetrads.

there is no dna synthesis between prophase I and the end of meiosis II, so you end up with the same amount of genetic material divided into 4 daughter cells.

thus the answer should be [b]e[/b][/hide]


linear algebra
matrix
vector
calculus
integration
function
derivative

Find the general solution of the system
\[ \begin{array}{l}
\dfrac{dx}{dt} \equal{} 5x \plus{} 3y \plus{} e^{2t}\\
\\
\dfrac{dy}{dt} \equal{} 2x \plus{} 2e^t\\
\\
\dfrac{dz}{dt} \equal{} x \plus{} y \plus{} e^t
\end{array}\]

I'd like to know the steps to solve them if possible? I understand that you have to use matrix methods or something

Let $ X$ be the vector $ \begin{pmatrix}x\\y\\z\end{pmatrix}$. The system can be written $ X'(t)\equal{}A\cdot X(t)\plus{}B(t)$, where $ A\equal{}\begin{pmatrix}5&3&0\\2&0&0\\1&1&0\end{pmatrix}$ and $ B(t)\equal{}\begin{pmatrix}e^{2t}\\2e^t\\e^t\end{pmatrix}$.
First, note that we don't need to know $ z$ to determine $ x$ or $ y$; we find those first, and then finding $ z$ is simply an integration problem. Let's ignore that variable for now, and just look at the first two. Let $ A_2$ be the upper left $ 2\times 2$ block of $ A$, $ X_2(t)$ be the vector of the first two functions, and $ B_2(t)$ the reduced version of $ B$.
The homogeneous solutions of $ X_2'(t)\equal{}A_2\cdot X_2'$ are of the form $ \exp(A_2 t)]\cdot U$ for vectors $ U$, so we would like to calculate that exponential. To do so, diagonalize: the trace of $ A_2$ is $ 5$ and its determinant is $ \minus{}6$, for eigenvalues of $ \minus{}1$ and $ 6$. We only really care about the column space of $ \exp(A_2 t)$, so we look for the eigenspaces: $ (A_2\plus{}I)\cdot \begin{pmatrix}1\\\minus{}2\end{pmatrix}\equal{}0$ and $ (A_2\minus{}6I)\cdot \begin{pmatrix}3\\1\end{pmatrix}\equal{}0$, so $ e^{\minus{}t}\begin{pmatrix}1\\\minus{}2\end{pmatrix}$ and $ e^{6t}\begin{pmatrix}3\\1\end{pmatrix}$ are a basis for the space of homogeneous solutions.
Next, we incorporate the non-homogeneous part, working with undetermined coefficients. Since neither $ e^{2t}$ nor $ e^t$ appears in the homogeneous solutions, there is a particular solution of the form $ e^t U\plus{}e^{2t} V$. Differentiate: $ e^t(A_2 U)\plus{}2e^{t}\cdot\begin{pmatrix}0\\1\end{pmatrix}\plus{}e^{2t}(A_2 V)\plus{}2e^{2t}\cdot\begin{pmatrix}1\\0\end{pmatrix}\equal{} e^t U \plus{}2e^{2t} V$. That splits into two equations, with solutions $ U\equal{}\begin{pmatrix}\minus{}\frac35\\ \frac45\end{pmatrix}$ and $ V\equal{}\begin{pmatrix}\minus{}\frac16\\\minus{}\frac16\end{pmatrix}$.

So, that leaves us with $ \begin{pmatrix}x(t)\\ y(t)\end{pmatrix}\equal{}c_1 e^{\minus{}t}\begin{pmatrix}1\\\minus{}2\end{pmatrix}\plus{}c_2 e^{6t}\begin{pmatrix}3\\1\end{pmatrix}\plus{}e^t \begin{pmatrix}\minus{}1\\ \frac53\end{pmatrix}\plus{}e^{2t} \begin{pmatrix}0\\\minus{}\frac23\end{pmatrix}$
Then, from $ z'(t)\equal{}x(t)\plus{}y(t)\plus{}e^t$, we get $ z(t)\equal{}c_1 e^{\minus{}t}\plus{}\frac{2}{3}c_2 e^{6t} \plus{}\frac65 e^t\minus{}\frac16 e^{2t}\plus{}c_3$.

[quote="jmerry"]Since neither $ e^{2t}$ nor $ e^t$ appears in the homogeneous solutions, there is a particular solution of the form $ e^t U \plus{} e^{2t} V$. Differentiate: $ e^t(A_2 U) \plus{} 2e^{t}\cdot\begin{pmatrix}0 \\
1\end{pmatrix} \plus{} e^{2t}(A_2 V) \plus{} 2e^{2t}\cdot\begin{pmatrix}1 \\
0\end{pmatrix} \equal{} e^t U \plus{} 2e^{2t} V$. That splits into two equations, with solutions $ U \equal{} \begin{pmatrix} \minus{} \frac35 \\
\frac45\end{pmatrix}$ and $ V \equal{} \begin{pmatrix} \minus{} \frac16 \\
\minus{} \frac16\end{pmatrix}$.[/quote]
Thanks, I get most of the solution now, except the part where you differentiate above - how do you get the LHS (with $ e^t(A_2U)$, etc)?

The left side is the derivative as determined by the differential equation. The right side is the derivative as determined by just doing it.


Pascal\u0027s Triangle
binomial coefficients
puzzles

.................g...h...i
..............d...e....f
...........a...b....c
........1
......1...1
....1...2...1

Above the topmost "1" of the Pascal's Triangle portion, there are three rows with 3 letters each in them.  They represent the the binomial coefficients for exponents of -1, -2, and -3,
respectively.  The adjacent numbers along the left side of the Pascal's Triangle, and the numbers that are adjacent to the "a",
"d", and "g" on their left-hand sides are all zeroes.

Determine the values of:  a,b,c,d,e,f,g,h, and i.


inequalities
trigonometry
blogs
inequalities proposed

[hide="This might help"] $x = \cos \alpha, y = \cos \beta, c = \cos \gamma$, where $\alpha+\beta+\gamma = \pi$.  ;) [/hide]

My solution:
The problem is equivalent to:
$ x, y, z>0,x^2+y^2+z^2+xyz=4\Rightarrow (x+y+z)^2\leq 8+xyz $
We have:
$ x, y, z>0,x^2+y^2+z^2+xyz=4\Leftrightarrow\sum\frac{x}{2x+yz}=1\Leftrightarrow $
$ \Leftrightarrow 1=\sum\frac{x^2}{2x^2+xyz}\Rightarrow 1\geq\frac{(x+y+z)^2}{2(x^2+y^2+z^2+3xyz} \Leftrightarrow $
$ \Leftrightarrow 2(x^2+y^2+z^2)+3xyz\geq (x+y+z)^2 \Leftrightarrow 2(4-xyz)+3xyz\geq (x+y+z)^2\Leftrightarrow $
$ \Leftrightarrow 8+xyz\geq (x+y+z)^2 $
_______________
Sandu Marin, Bucuresti, Romania
P.S. Salutari si urari de bine lui Virgil Nicula! Nu te-am uitat!
       Greetings and best wishes Virgil Nicula! I have not forgotten!

[u]Sa traiesti[/u], [b]Marinica[/b] ! [b]Ma bucur tare mult[/b]. [u]Ce fac cei de la[/u] [b]2512 & Co[/b] ?! Am vorbit cu [b]Simii[/b]. [u]See[/u] my [url=http://www.artofproblemsolving.com/Forum/blog.php?u=9882][color=red][b]blog[/b][/color][/url]. [u]Trecem pe[/u] [b]m.p.[/b]


It is just a convention.
By defining this, the combination formula $C_r^{n}=\frac{n!}{r!(n-r)!}$ is still valid for the case $n=r$.

[quote="ychjae"]我的一個白痴的理解方法：
明顯地，$n!=\frac{(n+1)!}{n+1}$
所以$0!=\frac{1!}{1}=1$ :blush:
這是從$x^{0}=x^{1-1}=\frac{x}{x}=1$的理解方法中想出來的 :oops:
(如有錯誤，請改正)[/quote]
我想應該可以這樣理解, 而且這想法亦並非白痴.

你跟我在发贴后不久想出的有些类似。
我的方法：
由定义：n！=n×（n-1）！
当n=1时，1！=1×0！
可得0！=1
 :lol: 另外：谢谢大家！:laugh:

顺便请教一下：如何在论坛中打公式？

I also have a stupid method which I came up with when I was young.
$(a+1)!-a!=a\cdot a!$
$1!-0!=0$
$0!=1$


find the minimum value of the function
$ f(x)\equal{}\sqrt{5a^2\plus{}4a\plus{}4}\plus{}\sqrt{5a^2\plus{}12a\plus{}8}$

it must be. Find the minimum value of the function f with the real number x
$ f(x)\equal{}\sqrt{5x^2\plus{}4x\plus{}4}\plus{}\sqrt{5x^2\plus{}12x\plus{}8}$

[quote="hoangvn.fix"]it must be. Find the minimum value of the function f with the real number x
$ f(x) \equal{} \sqrt {5x^2 \plus{} 4x \plus{} 4} \plus{} \sqrt {5x^2 \plus{} 12x \plus{} 8}$[/quote]
Let $ K\equal{}(0,x)$
$ \frac{f(x)}{\sqrt{5}}\equal{}\sqrt{(x\plus{}\frac{2}{5})^2\plus{}(\frac{4}{5})^2}\plus{}\sqrt{(x\plus{}\frac{6}{5})^2\plus{}(\frac{2}{5})^2}\equal{}OA\plus{}OB$ [geogebra]8fff3863d473b1654fcac319a9c93993d47a9798[/geogebra]

Clearly $ OA\plus{}OB$ is minimized when $ O$ is on the line $ AB$ , i.e. when $ OC\equal{}0.8 \times \frac{2}{3}\equal{}\frac{8}{15}$, i.e. when $ x\equal{}\minus{}\frac{14}{15}$

So min. of f $ \equal{}f(\minus{}\frac{14}{15})\equal{}\sqrt{\frac{52}{5}}$

hello, we get $ x_{min}\equal{}\minus{}\frac{14}{15}$ and $ y_{min}\equal{}\frac{2}{5}\sqrt{13}\sqrt{5}$
Sonnhard.


A two-digit number [i]n[/i] has the property that the sum of the digits of [i]n[/i] is the same as the sum of the digits of [i]6n[/i]. How many such numbers are there?

[quote="Brut3Forc3"][hide]We can tell that this is only possible when $ n$ is divisible by 9. Now it is easy to test all two-digit numbers divisible by 9. Testing gives the possible values of 18, 27, 36, 45, 90, and 99, so the answer is 6.[/hide][/quote]

The official answers says its 8.

Also, how can you "tell that this is only possible when n is divisible by 9"?

My bad, 54 and 72 also work. (I accidentally tested by multiplying by 9, not 6.) So it is in fact 8.
Okay, when divided by 3 or 9, the sum of digits of a number has the same remainder as the number itself does. So if n is not divisible by 3 or 9, then 6n will be, and then the sum of their digits cannot be the same.
If n is divisible by 3 but not by 9, then 6n will be divisible by 9, so then the sum of their digits cannot be the same.
Therefore, n must be divisible by 9.


A function $ f$ from the integers to the integers is defined as follows:
\[ f(n) \equal{} \begin{cases} n \plus{} 3 & \text{if n is odd} \\
n/2 & \text{if n is even} \end{cases}
\]Suppose $ k$ is odd and $ f(f(f(k))) \equal{} 27$. What is the sum of the digits of $ k$?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$

[hide]Work backwards on the problem.  From 27, it must go to 54 because 24 wouldn't go to 27.  From 54, there are 2 choices, 108 or 51.  Because 51 is odd, it will only lead to an even number for k.  From 108, subtract 3 to get 105, the odd k.  The sum of the digits of 105 is $1+0+5=6$, making the answer to the problem $\boxed{B}$[/hide]

You need to do it three times, not two. ;)

I still say he's right...
[hide]
Even though he forgot to iterate through $f(x)$ three times, i still believe he came to the right answer
if you just take his steps and complete the third iteration you get
$k = 105$
$1+0+5=6$
So the answer is $B$
[/hide]

Well, he got lucky. It's still better for him to do it the right way. ;)

Since $k$ is odd, then $f(k)=k+3$ which is even, so then the third would be $f(\frac{k+3}{2})$ and since that is also even, we have $\frac{k+3}{4}=27$ and so $k=105$ and the sum of the digits is $\boxed{6}$

[hide]Since 27 is odd, we see that $f(f(k))$ must be even, and thus must be 27*2=54. Assume that $f(k)=54\cdot2=108$. Then k must be 105 (instead of 216), since it is odd.

If $f(k)=54-3=51$, then k must be either 48 or 102, none of which are odd.

The sum of the digits of 105 is 6.  [/hide]


In a wargame, there are three kinds of units: artillery, cavalry, and infantry. Each turn, each side rolls two dice; the total on the dice is the number of "moves" that side has for that turn.

A certain player wants to use an artillery to kill as many infantry as possible (assume that there are 100 infantry or so, or enough to keep going as probability allows).  Both sides roll a die; if the side with the artillery wins the roll (rolls higher than his opponent), he kills an infantry AND can continue attacking without finishing that "move." Otherwise, his "move" is used up.

Derive a formula in terms of $m$, the number of moves, for the expected value of infantry killed in $m$ moves.

Eh? Arnav, do you have a solution, eh?

 :D

Or anyone else -- Berman?

For $m=1$, we have the expected value as being

\[\frac{7}{12}\left(\sum_{i=0}^{\infty}{i\left(\frac{5}{12}\right)^{i}}\right)\]

which is a [i]bit[/i] ugly...

[quote="not_trig"]
For $m=1$, we have the expected value as being
\[\frac{7}{12}\left(\sum_{i=0}^{\infty}{i\left(\frac{5}{12}\right)^{i}}\right) \]
which is a [i]bit[/i] ugly...[/quote]

Or 5/7 after simplifying.

I'm not sure that I'm following a couple of details.  My new interpretation based on not_trigs second post (below) is hidden below

[hide]Player A rolls two dice and gets a total of 4.  Now we have the following sequence of die tosses for Player A and Player B
A     B
4     3
3     1
2     1
5     5

Now 3 infantry are dead.  A's 1st move is over.  A now gets 3 more moves similar to this because he rolled a 4 to begin with.[/hide]

The probability that Player A beats Player B on a roll of one die is 5/12.

We're to find the expected number of infantry killed in m moves.    For m=1, I'm getting:

$\sum_{i=1}^{\infty}{\left(\frac{5}{12}\right)^{i}}= \frac{5}{7}$ as posted above.

The formula for the expected number of infantry killed in m moves seems fairly straighforward from here.

Although it isn't asked for, we should be able to find the expected number of infantry killed on one "turn"  where the number of moves per turn is determined by the first roll.  This might be fun.

I think I understand the problem now. (Edited based on not_trig's explanation below - I previously misread and thought both players were rolling 2 dice in "attack" mode.)

Hmm...

I think the problem was that I said that [i]each[/i] side rolls two dice... actually, Player A rolls 2 dice, and he gets that many moves.

The rest of the moves, each side rolls only $1$ die. So the chance that A beats B in one roll (both sides roll one) is $\frac{5}{12}$. Sorry for any confusion.

Oh yes, and all the rolls are for moves, not for the number of moves like you thought for the first roll. Since $m$ is already given, we know how many moves A will make.

Umm... so the expected value of men killed in one turn is 5/7 and the expected value of men killed in m turns is 5m/7... Am I confused? Actually, I can answer that-I am confused. A better question: am I missing something?

Hmm......

Wouldn't it be $\left(\frac{5}{7}\right)^{m}$?

I'm more confused by what you just said... :huh:

I was thinking along the same line as jb05.  Every move has the same expected number of dead infantry.  Since the moves are independent, just add the expected deaths from each move.

I can't see how the expected number of deaths could be less in m moves than for 1 move [(5/7)^m???].

Finding the expected number of deaths on one "turn" might be slightly more challenging.

Some rule changes might make the probabilities and expectations a bit more challenging.

Taking into account the first roll... the expected sum of the top faces of two dice is clearly 6, making m=6 and the total value 30/7 (so a little over 4).

If you don't believe that the expected value of m is 6, consider that for each way to get a combined value of 6-x (with rolls of a and b), there is exactly one way to get a combined value of 6+x (with rolls of 6-a and 6-b) unless a=b=3, but then the combined value is 6 anyway. So the average (expected value since there are equal probabilities for each instance) is 6.

If you don't believe that the expected value for m turns is 5m/7... the expected value for the first roll is 5/7. The expected value for the second roll is 5/7 since this is independent of the first roll (the number of soldiers killed on the first turn has no effect on the second turn). So on average, 5/7 of a soldier is killed, and then another 5/7 of a soldier is killed, so 10/7 of a solider is killed.

I hope that finishes off the problem.

OK, I get it now. Thanks  :D

I was also thinking about Risk: Attacker rolls 3 dice, defender 2. What is the Attacker's chance of winning two rolls, one roll, and 0? (Attacker's top 2 paired, in order, against defender's in order; defender wins ties)

[hide="If you do not play Risk"]
EX: Attacker - 5, 3, 2
Defender - 2, 4

Pairing:
A - D
5 - 4
3 - 2

A wins 2

Attacker - 2, 3, 3
Defender - 6, 3

A - D
3 - 6
3 - 3

D wins 2

Attacker - 4, 3, 6
Defender - 5, 4

A - D
6 - 5
4 - 4

Both win 1[/hide]

[quote="jb05"]Taking into account the first roll... the expected sum of the top faces of two dice is clearly 6, making m=6 and the total value 30/7 (so a little over 4). [/hide]

I thought that the expected sum of the top faces of two dice is 7...

Expected sum on one die is $\frac{1+2+3+4+5+6}6=3.5$, so the expected sum on two dice is $3.5\cdot2=7$.

...it was late, and my subconscious thought it was a 7-sided die... the same argument works though. But then the expected value is even easier--5!

In risk, I believe the chance that one attacker and one defender is lost is almost exactly 50% and the attacker is a little under 10% more likely than the defender to win both rolls. It's far too messy to write out, and I didn't do that correctly anyway, but I think it's more or less accurate. I'm not too fond of brute force probability.

[quote="not_trig"]Eh? Arnav, do you have a solution, eh?
[/quote]

LOL, I think Arnav has been...err..removed from this particular subforum.

[quote="not_trig"]Eh? Arnav, do you have a solution, eh?

 :D

Or anyone else -- Berman?

For $m=1$, we have the expected value as being
\[\frac{7}{12}\left(\sum_{i=0}^{\infty}{i\left(\frac{5}{12}\right)^{i}}\right) \]
which is a [i]bit[/i] ugly...[/quote]

Wait, I have sum of [b]i[/b](5/12)^i. Maybe it (my formula) is wrong?


For all $ a,b,c$ be nonnegative real numbers, prove that
\[ \frac{b^2\plus{}c^2}{a^2\plus{}bc}\plus{}\frac{c^2\plus{}a^2}{b^2\plus{}ca}\plus{}\frac{a^2\plus{}b^2}{c^2\plus{}ab}\ge\frac{2a}{b\plus{}c}\plus{}\frac{2b}{c\plus{}a}\plus{}\frac{2c}{a\plus{}b}\]


Hello.

Differential equation under exact equation:

Heres my work up to finding the integrating factor. Im using the same method I have seen from examples and the internet and it worked before on several exercises, but for this particular problem (and a few more) it just doesnt work out!

Take a look: after I find the integrating factor using the formula $ \frac {N' \minus{} M'}{M}$ the expression doesnt simplify! Is this method even correct?

[img]http://img107.imageshack.us/img107/4466/helppc0.jpg[/img]

Help! :)

We have $ \frac{dx}{dy}\equal{}\frac{3x^2\minus{}2xy^2}{y^4\minus{}4xy}$.

Used substitusion $ x(y)\equal{}y^3z(y)$.


linear algebra
matrix
linear algebra unsolved

Show that absolute value of eigenvalue of matrix is less than sum of absolute values of it's elements.

If $ A\equal{}(a_{i,j})$ and $ \lambda$ is an eigenvalue of $ A$ and $ x\equal{}(x_j)$, $ x \neq 0$, is its corresponding eigenvector, then we have $ Ax\equal{}\lambda x$. Thus we have

$ a_{i,1}x_1\plus{}a_{i,2}x_2\plus{}\cdots \plus{} a_{i,n}x_n\equal{}\lambda \cdot x_i$, where $ i\equal{}\overline{1,n}$, $ (*)$

Let $ k$ such that $ |x_k|\equal{} \max_{i\equal{}\overline{1,n}} |x_i|$.

From $ (*)$ we know that $ a_{k,1}x_1\plus{}a_{k,2}x_2\plus{}\cdots \plus{} a_{k,n}x_n\equal{}\lambda \cdot x_k$ ,equivalent to

$ |\lambda| \equal{} \left| a_{k,1} \cdot \frac{x_1}{x_k} \plus{} a_{k,2} \cdot \frac{x_2}{x_k} \plus{} \cdots \plus{} a_{k,n} \cdot \frac{x_n}{x_k} \right| \leq |a_{k,1}|\plus{}|a_{k,2}|\plus{}\cdots \plus{} |a_{k,n}|$, since $ \left| \frac{x_i}{x_k} \right| \leq 1$.And the conclusion follows.


Show that $ { 3m \choose 0} \plus{} { 3m \choose 1} \plus{} ... \plus{} { 3m \choose m\minus{}1}$ is asymptotically equal to $ { 3m \choose m}$.

An earlier part of the question suggested that the Stirling Approximation may be useful.


Find a partition of the real numbers set into two classes which have the same cardinality.
  Keywords: partition, cardinality.

I don't get it, man :?. Doesn't $(-\infty,0],(0,\infty)$ work? Am I missing something here?

Yes, you are missing. In fact, I am missing: the classes must be dense in the set of real numbers. Sorry  :(  :P

I'm sure a partition of $(0,1)$ instead of $\mathbb R$ will make all of us just as happy. :D

Write each number in $(0,1)$ as $0.a_1a_2a_3,\ldots$, where $a_i$ are binary digits. Put in a class of the partition those numbers for which there exists $k$ s.t. exactly one of $a_{2n+1},a_{2n+2}$ is a $1$, for all $n\ge k$, and put the rest of $(0,1)$ in the other class.

Does this construction work? It looks fine to me, but I'm in a hurry to write this, so I bet I'm doing something obviously wrong, but I'm sure you guys will find the power to forgive me if this is indeed the case. It's almost Christmas for crying out loud! :D:D

The answer  is trivial .  For ex : $Q$(rational numbers) and  $\{\sqrt{2}+Q\} $ are dense in  $R$ (real numbers) and $Q \cap \{\sqrt{2}+Q\} = \phi$
It sufficies now  to separe $R-Q-\{\sqrt{2}+Q\}$ in two arbitrary disjoint sets, $R_1,R_2$ , having the same  cardinal, and here you are  :
$C_1 = Q \cup R_1 , C_2 = \{\sqrt{2}+Q\}\cup R_2$ verify the initial conditions

 :cool: In the same way you can find a partition of $R$ into $ \aleph _0$(countable cardinal) classes such that each class is dense in $R$ and has  the same cardinal than$R$. :cool:


What is the numeral that will be listed directly beneath 25
when this triangular array is continued in the same manner?
\begin{tabular}{lllll}
1\\
2 & 3\\
4 & 5 & 6\\
7 & 8 & 9 & 10\\
11 & 12 & 13 & 14 & 15\\\end{tabular}

1
2   3
4   5   6
7   8   9   10
11 12  13  14  15
16 17  18  19  20  21
22 23  24  25

Find patterns to find 25 quickly. For example, 21 is a triangular number, and all of those are on the outside edge. 25 is 4 terms later, going in the 4th column.
Looking at 25's column it is +4, +5, +6, so the next term is 25 + 7= 32


Hello, I have a question about the basis, could anyone help?

Let $T$ be a linear operator on a finite dimensional vector space $V$

Prove that there is a basis $\{v_{1}, ..., v_{n}\}$ of $V$ such that $\{Tv_{1}, ..., Tv_{k}\}$ is a basis of im$T$ and $\{v_{k+1}, ..., v_{n}\}$ is a basis of ker$T$

take a basis for $kerT$, then extend it to a basis for $V$. the ones which arent part of the basis of $kerT$ , are the $v_{k}'s$ you are looking for


We will be holding a free Math Jam to discuss the problems and solutions of Round 1 of the USAMTS.  The Math Jam is on Thursday, October 11, at 7:30 PM Eastern / 4:30 PM Pacific.

[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Click here for more information about Math Jams.[/url]

I should add:

[b]Do not discuss Round 1 until the conclusion of the Math Jam.[/b]

This is so that we can contact students on Wednesday and/or Thursday AM who made a bona fide attempt to submit their solutions via email or fax on time, but because of some technical glitch (on our end) the solutions didn't get processed properly. If the fault was indeed on our end, or was something legitimately unforeseen by the student, then we'll give those students a chance to resubmit.

Note that the above paragraph does not cover "I tried to fax at 11:58 p.m. but got a busy signal" or "I tried to email at 11:59 p.m. but my internet went down." Do not wait until the last possible moment to submit your solutions.

Unfortunately, I can't come because I have a martial arts class exactly at 7:30 PM Eastern on Thursday :( ...will the (summary/extract/anything of) discussion be posted on the forum later?

If you go to the math jams section of the web site and click on the transcripts tab near the top of the page, you will find the transcripts to previous math jams listed in reverse chronological order.  The transcript for the USAMTS math jam will probably be posted there by Friday.

The [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=216]transcript[/url] from the Math Jam is now available.


probability
combinatorics unsolved
combinatorics

I need to know how to solve these questions please.. the methods to use are Probability and Combinations, you can use Hyper geometeric, Geometeric, Binomial and Uniform to solve these questions

1) Two Urns Contain Red and black balls, all alike except for colour. Urn A has 2 reds and 1 black, and Urn B has 101 reds ands 100 blacks. An urn is choosen at random, and you win a price if you correctly name the urn on the basis of the evidence of two balls drawn from it. After teh first ball is drawn and its colour reported, you can decide whether or not the ball shall be replaced before the second drawing. How do you order the second drawing and how do you decide on the urn?

2) King Asad, to test a candidate for the position of wise man, offers him a chance to marry the young lady in the court with the largest dowry. the amounts of the dowries are wirtten on slips of paper and mixed. A slip is drawn at random and the wise man must decide whether that is the largest dowry or not. if he decides it is, he gets the lady and her dowry if he is correct; otherwise he gets nothing. if he decides against the amount written on the first slip he must choose or refuse the next slip, and so until he chooses one or else the exhausted. In all, 100 attractive young ladies participate, each with a different dowry. How should the wise man make his decision?...
There is no real true solution but what appraoch would maximize the chances of success

3) There are fours cards in a box. One card is red on both sides. The other cards are each red on one side and green on the other. A card is drawn randomly from the box and placed on a table so that the down side cannot be seen. If the up side is red, what is the probability that the down side is also red ?

4) The rules of a game are as follows. A player (P) tosses a coin and the houses agrees to pay the player (.)(x) if P gets a head on the first toss, (.) (x)^2 if P gets a head on the second toss, (.)x^3 if P gets a head on the third toss, and so on. Find the mathematical expectation of P when
a) x = 1 b) x = 1.99 c) x=2
Write down any odd results that arise in the problem....

I would really greatly appreciate it that if you guys can solve it and show me how you guys did the problem as I have no clue what so ever to solve these questions ... Thank you... please show full method of solving the problem .. Bye... (I also need this asap)

Nebody ... please ? I really don't understand how to do these problems :| ... I mainly help with the "King Asad" one

Aren't you supposed to post each problem saparately?

I am too lazy to read all the questions. I will solve the "king Asad" one.

What he can do is to pass n slips. After that he can choose a slip that has bigger amount than the n slips.
Let P(n) be the probablity that he would be successful with this strategy. First the probablity that kth slip is the biggest is $\frac{1}{100}$. If this slip is in the n slips, he fail. To be successful, the biggest among the previous k-1 slips should be in the n slips. Therefore P(n) is $\frac{1}{100}*\frac{n}{n}+\frac{1}{100}*\frac{n}{n+1}+\frac{1}{100}*\frac{n}{n+2}+...+\frac{1}{100}*\frac{n}{99}$. This is maximum when n=37.

Since you did not understand what I said, I will try to explain each steps.

[quote="lightrhee"]Aren't you supposed to post each problem saparately?

I am too lazy to read all the questions. I will solve the "king Asad" one.[/quote]

I will get to the other problems once you are finished with this problem.

[quote="lightrhee"]What he can do is to pass n slips. After that he can choose a slip that has bigger amount than the n slips.[/quote]

Let n be any positve integer that is smaller than 100. He will just go through n slips and he is not going to pick any of the slips. After the n slips, he will pick a slip when he sees a slip with more dowry than any of the n slips.

[quote="lightrhee"]Let P(n) be the probablity that he would be successful with this strategy. First, the probablity that kth slip is the biggest is $\frac{1}{100}$.[/quote]

This must be clear. Let kth slips be the slip with the biggest dowry.

[quote="lightrhee"]If this slip is in the n slips, he fail. To be successful, the biggest among the previous k-1 slips should be in the n slips.[/quote]

To be successful,
1) the biggest dowry should not be in the first n slips(This means $n+1 \leq k$), and
2) the biggest dowry among the first k-1 slips should be in the first n slips.

[quote="lightrhee"]Therefore P(n) is $\frac{1}{100}*\frac{n}{n}+\frac{1}{100}*\frac{n}{n+1}+\frac{1}{100}*\frac{n}{n+2}+...+\frac{1}{100}*\frac{n}{99}$. This is maximum when n=37.[/quote]

The probability that,
1) the biggest dowry is the kth slip, and
2) the biggest dowry among the first k-1 slips is in the first n slips,
is  $\frac{1}{100}*\frac{n}{k-1}$. Add them up for all k such that $n+1 \leq k \leq 100$ gives us P(n).
To make the probabilty maximum, we need to find n that has the biggest P(n). It turns out that $n\approx \frac{100}{e} \approx 37$. The probability is about $\frac{1}{e}$.

You don't need to use PM. I often revisit my replies.
e is a non-algebraic number(By the way, this is no longer combinatorics). It can be defined in many ways.
One way would be
\[e=lim(1+\frac{1}{n})^n\]
e is about 2.71828. It has many nice properties. for example,
\[(e^x)'=e^x\]
\[e^{i \pi}+1=0\].

[quote="Biohazard Sniper"]is there ne way of solving it.. i cant have ne other way except for combinations and permutations[/quote]
I don't understand what you are saying. Must be my English(I came to Canada 3 years ago. I'm still learning English.)

[quote="Biohazard Sniper"]i can only use permuations or combinates.. and that would be the basics ... like hyper geometeric, geometeric, uniform and bionomial. i cant use ne other way[/quote]
hmm. This is only solution I have. :(  I will think about another way.

[quote="Biohazard Sniper"]
3) There are fours cards in a box. One card is red on both sides. The other cards are each red on one side and green on the other. A card is drawn randomly from the box and placed on a table so that the down side cannot be seen. If the up side is red, what is the probability that the down side is also red ? [/quote]

There are 5 red sides. Three of them has green on the other side. Therefore, the probability is $\frac{2}{5}$.

oh i got that one, i was talking about the kign asad one, that is the one badly need

[quote="Biohazard Sniper"]i need it in an hour at the most if that is possible please[/quote]
An hour! I don't think I can come up with a new solution within an hour. :(  Unless, other users help. I'm really sorry. Why do you need this in an hour? :?

please, i woudl be so grateful to you, all you have to do is make up ne method that would possibly work to solve that question

[quote="Biohazard Sniper"]because i have a test tommorow and i need to understand how do such a question on the test, otherwise i won't be able to two to three of the questions related to this one ... so yea[/quote]
Where are you writing this test? I don't think this is school test(I'm a high school student. High school tests are very different from this kind of question. :D )

[quote="Biohazard Sniper"]this is high school, grade 12 university enhanced.. crazy math that i don't understand yet i took... i just need a possible solution to the question king asad ... if you may ... please[/quote]
Again, I must say sorry. This is my limit. I can not think any other approch. This is high school test? Wow!!!!!!!!!! I am sure that everyone would fail. I was last year's IMO team member. This means I'm the one of the best 6 students in Canada at math, but even I can not solve this. So I seriously do not think you are suppose to know this...

[quote="Biohazard Sniper"]do you know a method of approaching this question, don't solve it ... just a method you might take[/quote]
Once you solve a problem, it is hard to find a new solution. The perspective doesn't easily change(This is my belief).
I will not be able to come up with another method in an hour(I'm also somewhat sleepy :lol: ). Maybe in a few days, but not in an hour.
I am still not convinced that this is a high school problem. :D Why won't they let you use some algebra? This is evil. :D  :D