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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 2016, when a=2 and b=671. 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 \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.
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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 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.
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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>.
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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> since <math>503</math> is prime. 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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===Shortcut===
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After deducing that <math>2012</math> and case <math>1</math> is impossible, and since there is no option for <math>0</math>, <math>2016</math> is obviously a solution and the answer is <math>\boxed{\textbf{(A)}\ 1}</math>.
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-mathboy282
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==Video Solution by Pi Academy==
 +
 
 +
https://youtu.be/6Pgo0Tvp2Sk?si=2XeBRkt9Xe_QB7v3
 +
 
 +
~ Pi Academy
 +
 
 +
==Video Solution 2==
 +
https://youtu.be/7QNI7VMFkow
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution 3 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=WzSddwBAWtA
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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]]
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{{MAA Notice}}

Latest revision as of 21:25, 10 October 2024

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 $4$ since $503$ is prime. 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 $\boxed{\textbf{(A)}\ 1}$.

Shortcut

After deducing that $2012$ and case $1$ is impossible, and since there is no option for $0$, $2016$ is obviously a solution and the answer is $\boxed{\textbf{(A)}\ 1}$.

-mathboy282

Video Solution by Pi Academy

https://youtu.be/6Pgo0Tvp2Sk?si=2XeBRkt9Xe_QB7v3

~ Pi Academy

Video Solution 2

https://youtu.be/7QNI7VMFkow

~savannahsolver

Video Solution 3 by SpreadTheMathLove

https://www.youtube.com/watch?v=WzSddwBAWtA

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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