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

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==Solution==
 
==Solution==
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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A positive integer with only four positive divisors has its prime factorization in the form of <math>a \cdot b</math>, where <math>a</math> and <math>b</math> are both prime positive integers or <math>c^3</math> where <math>c</math> is a prime. One can easily deduce that none of the numbers are even near a cube so the second case is not possible. We now look at the case of <math>a \cdot 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.  
  
 
Case 1: Either <math>a</math> or <math>b</math> is 2.  
 
Case 1: Either <math>a</math> or <math>b</math> is 2.  
  
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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If this is true then we have to have that one of <math>(a+1)</math> or <math>(b+1)</math> 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 <math>\frac{2016}{3} - 1</math>  or <math>\frac{2010}{3} - 1</math> is a prime. We see that in this case none of them work.
  
 
Case 2: Both <math>a</math> and <math>b</math> are odd primes.  
 
Case 2: Both <math>a</math> and <math>b</math> are odd primes.  
  
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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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>.
<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. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math> as <math>2016</math> is the only solution.
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<math>2012={4}\cdot{503}</math> so either <math>(a+1)</math> or <math>(b+1)</math> both have a factor of <math>2</math> or one has a factor of <math>4</math>. If it was the first case, then <math>a</math> or <math>b</math> will equal <math>1</math>. That means that either <math>(a+1)</math> or <math>(b+1)</math> has a factor of <math>4</math>.  That means that <math>a</math> or <math>b</math> is <math>502</math> which isn't a prime, so 2012 does not work.  <math>2016 = 4 \cdot 504</math> so we have <math>(503 + 1)(3 + 1)</math>. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.
  
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== See also ==
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{{AMC10 box|year=2013|ab=B|num-b=23|num-a=25}}
  
 
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[[Category:Introductory Number Theory Problems]]
**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?**
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{{MAA Notice}}

Revision as of 22:48, 17 May 2021

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 $a \cdot b$, 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 the second case is not possible. We now look at the case of $a \cdot b$. 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 in this case none of them work.

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 $2016$. $2012={4}\cdot{503}$ so either $(a+1)$ or $(b+1)$ both have a factor of $2$ or one has a factor of $4$. If it was the first case, then $a$ or $b$ will equal $1$. That means that either $(a+1)$ or $(b+1)$ has a factor of $4$. That means that $a$ or $b$ is $502$ which isn't a prime, so 2012 does not work. $2016 = 4 \cdot 504$ so we have $(503 + 1)(3 + 1)$. 503 and 3 are both odd primes, so 2016 is a solution. Thus the answer is $\boxed{\textbf{(A)}\ 1}$.

See also

2013 AMC 10B (ProblemsAnswer KeyResources)
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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