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

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==Solution==
 
==Solution==
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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=7</math> and <math>b=251</math>. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.
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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 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 <math>a*b</math>. 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>. Easily we can see that Now we can take cases again.
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Case 1: Either <math>a</math> or <math>b</math> is 2.
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If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> 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 <math>\frac{2016}{3} - 1</math> or <math>\frac{2010}{3} - 1</math> is a prime. We see that <math>2016</math> is the only one that works in this case.
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Case 2: Both <math>a</math> and <math>b</math> are odd primes.
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This implies that both <math>(a+1)</math> and <math>(b+1)</math> are even which implies that in this case the number must be divisible by <math>4</math>. This leaves only <math>2012</math> and <math>2016</math>. We know <math>2016</math> works so it suffices to check whether <math>2012</math> works.
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<math>2012=4*503</math> so we have that a factor of <math>2</math> must go to both <math>(a+1)</math> and <math>(b+1)</math>. So we have that <math>(a+1)</math> and <math>(b+1)</math> equal the numbers <math>(2+503)(2+1)</math>, but this contradicts our assumption for the case.
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Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math> as <math>2016</math> is the only solution..

Revision as of 14:25, 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 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 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..
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