Talk:2012 USAMO Problems/Problem 3

Revision as of 16:14, 3 May 2012 by Lightest (talk | contribs)

The answer is the set of all integers that are at least 3.

For composite n where there are two primes p_1 and p_2 such that n/2<p_1<p_2<n, here's your construction:

Pick maximal integers j_1 and j_2 such that$((p_1)^(j_1))((p_2)^(j_2))$ divides i.

Pick a minimal positive integer s such that $(n(n+1)/2)+(s-1)(p_1)$ is 0 mod p_2. (You know it exists since p_1 and p_2 are relatively prime.)

Pick an integer t such that$(n(n+1)/2)+(s-1)(p_1)+(t-1)(p_2)=0$. (It exists because of how we defined s. It also must be negative.)

Then $a_i=(s^(j_1))(t^(j_2))$.

For n=4:

$a_i=(-1)^(j_1+j_2)$, where$(2^j_1)(3^j_2)$divides i.

For n=6:

$a_i=(2^j_1)(-5)^j_2$, where $(3^j_1)(5^j_2)$divides i.

For n=10:

$a_i=(2^j_1)(-9)^j_2$, where $(5^j_1)(7^j_2)$divides i.

[I don't know LaTeX, so someone else can input it.]

--Mage24365 09:00, 25 April 2012 (EDT)