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"

(→‎Solution)
(→‎Solution)
Line 7: Line 7:
  
 
==Solution==
 
==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. 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).  Since all odd numbers have an odd number in their prime factorization, the product of (a+1)(b+1) results in an even number.  Therefore we can rule all out all of the odd integers in the set.  Considering the even cases, we see that 2016 is the only one that works. Thus the answer is (A) 1.
+
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. 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).  Since all odd numbers have an odd number in their prime factorization, the product of (a+1)(b+1) results in an even number.  Therefore we can rule all out all of the odd integers in the set.  Considering the even cases, we see that 2016 is the only one that works. Thus the answer is <math>\textbf{(A)}\ 1</math>.

Revision as of 17:35, 21 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 nice?


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

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. 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). Since all odd numbers have an odd number in their prime factorization, the product of (a+1)(b+1) results in an even number. Therefore we can rule all out all of the odd integers in the set. Considering the even cases, we see that 2016 is the only one that works. Thus the answer is $\textbf{(A)}\ 1$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_24&oldid=51267"