AMTI Practice corner !

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 .

AMTI, RMO, NMTC

Note by Aditya Narayan Sharma
3 years, 7 months ago

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1 vote

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Problem 2 :

In=cos(nθ)cosθI_n = \int \frac{cos(n\theta)}{cos\theta}

Show that : (n1)(In+In2)=2sin(n1)θ\text{Show that : } (n-1)(I_n+I_{n-2}) = 2sin(n-1)\theta

Aditya Narayan Sharma - 3 years, 7 months ago

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Problem 1 :

Prove that there are infinitely many reals pp such that p(p3p)p(p-3\text{{p}}) is an integer where p is not an integer.

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

Aditya Narayan Sharma - 3 years, 7 months ago

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Let f(x)=x23x{x} f(x) = x^2 - 3x\{x\} .

Consider the interval [n,n+1] [n, n + 1] for integer n.
f(n)=n2 f(n) = n^2
f(n+1)=(n+1)2f(n + 1) = (n + 1)^2

For p[n,n+1],f(p)=p23pn p \in [n, n + 1], f(p) = p^2 - 3pn is continuous, and hence takes every value between f(n) f(n) and f(n+1)f(n+1) , by the intermediate value theorem.

But we can then find p p such that f(p)=n2+1 f(p) = n^2 + 1 or whatever integer between n2 n^2 and (n+1)2 (n + 1)^2 .
And since there are infinitely many n n , we are done.

Ameya Daigavane - 3 years, 7 months ago

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

f(p)=p(p3p)=p([p]2p)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]I    2pI[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 xyz....5ˉ10=p\frac{\bar{xyz....5}}{10} = p as it's non-integer part is 0.5.

so f(p) = xyz....5ˉ10([p]1)    10([p]1)\frac{\bar{xyz....5}}{10}([p]-1)\implies 10|([p]-1)

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

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

Aditya Narayan Sharma - 3 years, 7 months ago

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@Aditya Narayan Sharma You may post the next problem.

Aditya Narayan Sharma - 3 years, 7 months ago

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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)?

A Former Brilliant Member - 3 years, 7 months ago

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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

Aditya Narayan Sharma - 3 years, 7 months ago

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thank you!

A Former Brilliant Member - 3 years, 7 months ago

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@A Former Brilliant Member Oh bdw check my note of online junior mathematician search. It's an o online contest. Lasts til 10th may. Join it plz

Aditya Narayan Sharma - 3 years, 7 months ago

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@Aditya Narayan Sharma I think that I already submitted the answers. I didn't get any message after the submission though.

A Former Brilliant Member - 3 years, 7 months ago

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@A Former Brilliant Member Oh when diu submit? Was it shown that your response has been recorded?

Aditya Narayan Sharma - 3 years, 7 months ago

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@Aditya Narayan Sharma Bdhoi amra pathaini ..otai bl6a

Sayandeep Ghosh - 3 years, 7 months ago

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@A Former Brilliant Member Yeah it must give u

Sayandeep Ghosh - 3 years, 7 months ago

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