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

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**Note:  <math>\frac{2016}{3} - 1</math> = <math>672</math> - <math>1</math> = <math>671</math> = <math>61</math> * <math>11</math> , so why did we get the right answer?**
 
**Note:  <math>\frac{2016}{3} - 1</math> = <math>672</math> - <math>1</math> = <math>671</math> = <math>61</math> * <math>11</math> , so why did we get the right answer?**
  
−
**Note: In Case 1, isn't one of <math>(a+1)</math> or $(b+1) even and one is 3, as one of a and b is 2 and one is an odd prime?**
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**Note: In Case 1, isn't one of <math>(a+1)</math> or <math>(b+1)</math> even and one is 3, as one of a and b is 2 and one is an odd prime?**
  
 
**Note: 2016 actually does not work in the first case, but it does work in the second case, so it is indeed the only solution.**
 
**Note: 2016 actually does not work in the first case, but it does work in the second case, so it is indeed the only solution.**

Revision as of 19:23, 15 November 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 odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even 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.


    • Note: $\frac{2016}{3} - 1$ = $672$ - $1$ = $671$ = $61$ * $11$ , so why did we get the right answer?**
    • Note: In Case 1, isn't one of $(a+1)$ or $(b+1)$ even and one is 3, as one of a and b is 2 and one is an odd prime?**
    • Note: 2016 actually does not work in the first case, but it does work in the second case, so it is indeed the only solution.**

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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