Difference between revisions of "Northeastern WOOTers Mock AIME I Problems/Problem 15"
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| − | We claim that <math>\phi(n)>n-\sqrt{n}</math> if and only if <math>n</math> is prime or | + | 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 21:33, 7 August 2021
Problem 15
Find the sum of all integers 
such that ![\[\phi(n)>n-\sqrt{n},\]](http://latex.artofproblemsolving.com/2/2/7/2278d67dcf2c3f9a73967961071ed4672ca776ae.png)
where 
denotes the number of integers less than or equal to 
that are relatively prime to .
Solution
We claim that 
if and only if is prime or 1.
IF: If is prime, then 
, which is true for all 
. For 
the statement is true as well.
ONLY IF: If is not prime and not 
, then must have a prime divisor 
such that 
; if this was not the case, then the number of not necessarily distinct prime factors could have would be , contradiction. It follows that
![\[\phi(n) \leq \left( 1 - \frac{1}{p} \right) \cdot n \leq \left( 1 - \frac{1}{\sqrt{n}} \right) \cdot n = n - \sqrt{n}.\]](http://latex.artofproblemsolving.com/a/b/a/aba2ffc0b37409be8d1665c00ce51ff11ecd68fb.png)
This proves both directions of the claim. All that remains is to sum the first 
prime numbers, starting with 
and ending with 
, and add to obtain an answer of 
.
