Difference between revisions of "Northeastern WOOTers Mock AIME I Problems/Problem 15"

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== Solution ==
 
== Solution ==
  
We claim that <math>\phi(n)>n-\sqrt{n}</math> if and only if <math>n</math> is prime or <math>1</math>.  
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We claim that <math>\phi(n)>n-\sqrt{n}</math> if and only if <math>n</math> is prime or 1.  
  
 
'''IF:''' If <math>n</math> is prime, then <math>\phi(n) = n - 1 > n - \sqrt{n}</math>, which is true for all <math>n \geq 2</math>. For <math>n = 1</math> the statement is true as well.
 
'''IF:''' If <math>n</math> is prime, then <math>\phi(n) = n - 1 > n - \sqrt{n}</math>, which is true for all <math>n \geq 2</math>. For <math>n = 1</math> the statement is true as well.

Revision as of 20:33, 7 August 2021

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 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 $89$, and add $1$ to obtain an answer of $\boxed{964}$.