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 $\textbf{(A)} | + | 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)} |
Revision as of 17:34, 21 February 2013
Problem
A positive integer 
is nice if there is a positive integer 
with exactly four positive divisors (including 
and ) such that the sum of the four divisors is equal to . How many numbers in the set 
are nice?
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)}
