Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
 
(5 intermediate revisions by the same user not shown)
Line 8: Line 8:
 
== Solution ==
 
== Solution ==
  
−
We claim that <math>\phi(n)>n-\sqrt{n}</math> if and only if <math>n</math> is prime.  
+
We claim that <math>\phi(n) > n - \sqrt{n}</math> if and only if <math>n</math> is prime or <math>1</math>.  
  
−
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>.  
+
'''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.
  
−
ONLY IF: If <math>n</math> is not prime, then <math>n</math> must have a prime divisor <math>p</math> such that <math>p \leq \sqrt{n}</math>; if this was not the case, then the number of not necessarily distinct prime factors <math>n</math> could have would be <math>1</math>, contradiction. It follows that
+
'''ONLY IF:''' If <math>n</math> is not prime and not <math>1</math>, then <math>n</math> must have a prime divisor <math>p</math> such that <math>p \leq \sqrt{n}</math>; if this was not the case, then the number of not necessarily distinct prime factors <math>n</math> could have must be <math>1</math>, contradiction. It follows that
−
<cmath> \phi(n) \leq \left( 1 - \frac{1}{p} \right) \cdot n \leq \left( 1 - \frac{1}{\sqrt{n}} \right) \cdot n = n - \sqrt{n} </cmath>.
+
<cmath> \phi(n) \leq \left( 1 - \frac{1}{p} \right) \cdot n \leq \left( 1 - \frac{1}{\sqrt{n}} \right) \cdot n = n - \sqrt{n}. </cmath>
  
−
This proves both directions of the claim. All that remains is to sum the first <math>24</math> prime numbers, starting with <math>2</math> and ending with <math>89</math>, to obtain an answer of \fbox{<math>963</math>}.
+
This proves both directions of the claim. All that remains is to sum the first <math>24</math> prime numbers and add <math>1</math> to obtain an answer of <math>\boxed{964}</math>.

Latest revision as of 21:35, 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 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 $\boxed{964}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Northeastern_WOOTers_Mock_AIME_I_Problems/Problem_15&oldid=159736"