# Northeastern WOOTers Mock AIME I Problems/Problem 15

## 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 or $1$.

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

ONLY IF: If $n$ is not prime and not $1$, 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 must 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 and add $1$ to obtain an answer of $\boxed{964}$.