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 "2013 AMC 10B Problems/Problem 24"
(→‎Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are nice?
+
A positive integer <math>n</math> is ''nice'' if there is a positive integer <math>m</math> with exactly four positive divisors (including <math>1</math> and <math>m</math>) such that the sum of the four divisors is equal to <math>n</math>. How many numbers in the set <math>\{ 2010,2011,2012,\dotsc,2019 \}</math> are mean and bullying other numbers?
  
  

Revision as of 16:31, 22 February 2013

Problem

A positive integer $n$ is nice if there is a positive integer $m$ with exactly four positive divisors (including $1$ and $m$) such that the sum of the four divisors is equal to $n$. How many numbers in the set $\{ 2010,2011,2012,\dotsc,2019 \}$ are mean and bullying other numbers?


$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5$

Solution

A positive integer with only four positive divisors has its prime factorization in the form of $a*b$, where $a$ and $b$ are both prime positive integers or c^3 where c is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of $a*b$. The four factors of this number would be $1$, $a$, $b$, and $ab$. The sum of these would be $ab+a+b+1$, which can be factored into the form $(a+1)(b+1)$. Easily we can see that Now we can take cases again.

Case 1: Either $a$ or $b$ is 2.

If this is true then we have to have that one of $(a+1)$ or $(b+1)$ is even and that one is 3. So we have that in this case the only numbers that work are odd multiples of 3 which are 2010 and 2016. So we just have to check if either $\frac{2016}{3} - 1$ or $\frac{2010}{3} - 1$ is a prime. We see that $2016$ is the only one that works in this case.

Case 2: Both $a$ and $b$ are odd primes.

This implies that both $(a+1)$ and $(b+1)$ are even which implies that in this case the number must be divisible by $4$. This leaves only $2012$ and $2016$. We know $2016$ works so it suffices to check whether $2012$ works. $2012=4*503$ so we have that a factor of $2$ must go to both $(a+1)$ and $(b+1)$. So we have that $(a+1)$ and $(b+1)$ equal the numbers $(2+503)(2+1)$, but this contradicts our assumption for the case. Thus the answer is $\boxed{\textbf{(A)}\ 1}$ as $2016$ is the only solution.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&oldid=51408"