Find the largest positive integer $n>10$ such that the residue of $n$ when divided by each perfect square between $2$ and $\dfrac n2$ is an odd number.
Problem
Source: Romanian JBMO TST 2005 - day 1, problem 2
Tags: inequalities, number theory proposed, number theory
Myth
02.04.2005 18:40
Ops, $n/2$!
wellknown
09.04.2005 22:06
115?
Valentin Vornicu
10.04.2005 17:25
This thread is locked until 24th of April.
Myth
25.04.2005 14:31
I think we can unblock this topic. My students got and prooved that answer is 505. It is much greater than 115. Hope they are right.
Valentin Vornicu
25.04.2005 14:34
They are right Mikhail!
perfect_radio
26.04.2005 00:34
i'm very interested to see the proof. by using the computer i also got the same result. this problem seems to be different from the one given ml contest i hope that i solved that one correctly, but i can't do this one
Valentin Vornicu
26.04.2005 01:57
It is different from the one on the ML contest. This one is much easier. You look for odd perfect squares between $\dfrac n4$ and $\dfrac n3$. If such a square say $k^2$ exists, then $ n= 3k^2 + r$ and as $n$ is odd, it's obvious that $r$ is even and we are done. Obviously for sufficiently large $n$ there exists an odd square between $\dfrac n4$ and $\dfrac n3$. Using some algebra and inequalities we can estimate such an $n$, and then we find close to him that $n=505$ verifies, thus it is the largest one.