I have heard about NMTC and AMTI , and this year I will be giving an olympiad for the first time . So let's have a AMTI practice corner, as the NMTC is 2 or 3 months from now.

Please contribute as many problems and their various solutions so as to help every other aspirant !!!

I am posting the first problem here .

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TopNewestProblem 2 :

$I_n = \int \frac{cos(n\theta)}{cos\theta}$

$\text{Show that : } (n-1)(I_n+I_{n-2}) = 2sin(n-1)\theta$

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Great initiative! Do you know how to register for the exam and what are the list of test centers(I am from Bangalore)?

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Yeah. NMTC is the first step. Visit amtionline.com and check the toll free number there . the forms are available at the first week of June. It's an all India test and the centres will definitely include Bangalore

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thank you!

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Problem 1 :

Prove that there are infinitely many reals $p$ such that $p(p-3\text{{p}})$ is an integer where p is not an integer.

$\mathbf{Clarification : } \text{ {.} denotes fractional part}$

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Let $f(x) = x^2 - 3x\{x\}$.

Consider the interval $[n, n + 1]$ for integer n.

$f(n) = n^2$

$f(n + 1) = (n + 1)^2$

For $p \in [n, n + 1], f(p) = p^2 - 3pn$ is continuous, and hence takes every value between $f(n)$ and $f(n+1)$, by the intermediate value theorem.

But we can then find $p$ such that $f(p) = n^2 + 1$ or whatever integer between $n^2$ and $(n + 1)^2$.

And since there are infinitely many $n$, we are done.

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Absolutely ! this is the standard approach. Let me just share another view which specifies the integers also.

$f(p) = p(p-3\text{{p}}) = p([p] - 2\text{{p}})$

Since f(p) is an integer and product of two factors wher p isn't a integer. So we must have ([p] - 2{p}) a integer as two non-integers never multiply to be a integer.

$[p] \in \mathbb{I} \implies \text{2{p}} \in \mathbb{I}$

It's clear that {p} = 0.5 for 2{p} to be a integer .

2{p} = 1 .

So p = [p] + 0.5. we may write p as $\frac{\bar{xyz....5}}{10} = p$ as it's non-integer part is 0.5.

so f(p) = $\frac{\bar{xyz....5}}{10}([p]-1)\implies 10|([p]-1)$

So $[p] = 10k+1$ $k \in [0,\infty)$

So we have infinitely many integers p of the form $(p = [p] + 0.5 = 10k+1.5)$

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