Northeastern WOOTers Mock AIME I Problems/Problem 15

Revision as of 21:23, 7 August 2021 by Skyguy88 (talk | contribs) (Solution)

Problem 15

Find the sum of all integers $n\le96$ such that \[\phi(n)>n-\sqrt{n},\] where $\phi(n)$ denotes the number of integers less than or equal to $n$ that are relatively prime to $n$.



Solution

We claim that $\phi(n)>n-\sqrt{n}$ if and only if $n$ is prime.

IF: If $n$ is prime, then $\phi(n) = n - 1 > n - \sqrt{n}$, which is true for all $n \geq 2$.

ONLY IF: If $n$ is not prime, then $n$ must have a prime divisor $p$ such that $p \leq \sqrt{n}$; if this was not the case, then the number of not necessarily distinct prime factors $n$ could have would be $1$, contradiction. It follows that \[\phi(n) \leq \left( 1 - \frac{1}{p} \right) \cdot n \leq \left( 1 - \frac{1}{\sqrt{n}} \right) \cdot n = n - \sqrt{n}\].

This proves both directions of the claim. All that remains is to sum the first $24$ prime numbers, starting with $2$ and ending with $89$, to obtain an answer of \fbox{$963$}.