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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Problem 2 :
In=∫cosθcos(nθ)
Show that : (n−1)(In+In−2)=2sin(n−1)θ
Problem 1 :
Prove that there are infinitely many reals p such that p(p−3p) is an integer where p is not an integer.
Clarification: . denotes fractional part
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Let f(x)=x2−3x{x}.
Consider the interval [n,n+1] for integer n.
f(n)=n2
f(n+1)=(n+1)2
For p∈[n,n+1],f(p)=p2−3pn 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)=n2+1 or whatever integer between n2 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−3p)=p([p]−2p)
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⟹2p∈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 10xyz....5ˉ=p as it's non-integer part is 0.5.
so f(p) = 10xyz....5ˉ([p]−1)⟹10∣([p]−1)
So [p]=10k+1 k∈[0,∞)
So we have infinitely many integers p of the form (p=[p]+0.5=10k+1.5)
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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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