# Difference between revisions of "2013 AMC 10B Problems/Problem 24"

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− | A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Since all odd numbers have an odd number in their prime factorization, the product of <math>(a+1)(b+1)</math> 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 the only one that works is <math>2016</math>, when <math>a= | + | A positive integer with only four positive divisors has its prime factorization in the form of <math>a*b</math>, where <math>a</math> and <math>b</math> are both prime positive integers. The four factors of this number would be <math>1</math>, <math>a</math>, <math>b</math>, and <math>ab</math>. The sum of these would be <math>ab+a+b+1</math>, which can be factored into the form <math>(a+1)(b+1)</math>. Since all odd numbers have an odd number in their prime factorization, the product of <math>(a+1)(b+1)</math> 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 the only one that works is <math>2016</math>, when <math>a=7</math> and <math>b=251</math>. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>. |

## Revision as of 11:31, 22 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 , where and are both prime positive integers. The four factors of this number would be , , , and . The sum of these would be , which can be factored into the form . Since all odd numbers have an odd number in their prime factorization, the product of 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 the only one that works is , when and . Thus the answer is .