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Difference between revisions of "2005 Alabama ARML TST Problems/Problem 13"

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For the first one, the sum of the reciprocals of the divisors of <math>n</math> is therefore <math>1+\dfrac{1}{p_1}+\dfrac{1}{p_1^2}+\dfrac{1}{p_1^3}+\dfrac{1}{p_1^4}+\dfrac{1}{p_1^5}</math>. The smallest prime (2) makes that less than 2, and if <math>p_1</math> gets bigger, then that expression gets smaller, so there is absolutely no way that <math>n=p_1^5</math>. So the second case is true.
 
For the first one, the sum of the reciprocals of the divisors of <math>n</math> is therefore <math>1+\dfrac{1}{p_1}+\dfrac{1}{p_1^2}+\dfrac{1}{p_1^3}+\dfrac{1}{p_1^4}+\dfrac{1}{p_1^5}</math>. The smallest prime (2) makes that less than 2, and if <math>p_1</math> gets bigger, then that expression gets smaller, so there is absolutely no way that <math>n=p_1^5</math>. So the second case is true.
  
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<center><math>\begin{eqnarray}1+\dfrac{1}{p_1}+\dfrac{1}{p_1p_2}+\dfrac{1}{p_1p_2^2}+\dfrac{1}{p_2}+\dfrac{1}{p_2^2}=2\\
+
<cmath>\begin{eqnarray*}1+\dfrac{1}{p_1}+\dfrac{1}{p_1p_2}+\dfrac{1}{p_1p_2^2}+\dfrac{1}{p_2}+\dfrac{1}{p_2^2}=2\\
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\dfrac{p_1p_2^2+p_1p_2+p_1+p_2^2+p_2+1}{p_1p_2^2}=\dfrac{(p^2+p_2+1)(p_1+1)}{p_1p_2^2}=2\\
+
\dfrac{p_1p_2^2+p_1p_2+p_1+p_2^2+p_2+1}{p_1p_2^2}=2\\
 
p_1p_2^2-p_1p_2-p_1=p_2^2+p_2+1\\
 
p_1p_2^2-p_1p_2-p_1=p_2^2+p_2+1\\
 
p_1(p_2^2-p_2-1)=p_2^2+p_2+1
 
p_1(p_2^2-p_2-1)=p_2^2+p_2+1
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\end{eqnarray}</math></center>
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\end{eqnarray*}</cmath>
  
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Therefore, <math>-p_1\equiv 1\bmod{p_2}</math>. Now the only way that that is possible is when <math>p_2=2</math>. Solving for <math>p_1</math>, we get that <math>p_1=7</math>. Checking, the sum of the reciprocals of the divisors of <math>\boxed{28}</math> indeed sum to 2, and 28 does have 6 factors.
+
Therefore, <math>p_1\equiv -1\bmod{p_2}</math>. Now the only way that that is possible is when <math>p_2=2</math>. Solving for <math>p_1</math>, we get that <math>p_1=7</math>. Checking, the sum of the reciprocals of the divisors of <math>\boxed{28}</math> indeed sum to 2, and 28 does have 6 factors.
 +
 
 +
Furthermore, if we rearrange <math>\textbf{(2)}</math>, we can see that the number would be the sum of all its divisors other than itself, making it a perfect number. Checking the perfect numbers, we find that the second perfect number, <math>28</math>, fits the required form and is indeed the solution.
  
 
==See also==
 
==See also==
 +
{{ARML box|year=2005|state=Alabama|num-b=12|num-a=14}}
 +
 +
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 18:14, 10 March 2015

Problem

There is one natural number with exactly 6 positive divisors, the sum of whose reciprocals is 2. Find that natural number.

Solution

Let the number be $n$, and let $p_1$ and $p_2$ be primes. Therefore, one of the following is true:

  • $n=p_1^5$
  • $n=p_1p_2^2$

For the first one, the sum of the reciprocals of the divisors of $n$ is therefore $1+\dfrac{1}{p_1}+\dfrac{1}{p_1^2}+\dfrac{1}{p_1^3}+\dfrac{1}{p_1^4}+\dfrac{1}{p_1^5}$. The smallest prime (2) makes that less than 2, and if $p_1$ gets bigger, then that expression gets smaller, so there is absolutely no way that $n=p_1^5$. So the second case is true.

\begin{eqnarray*}1+\dfrac{1}{p_1}+\dfrac{1}{p_1p_2}+\dfrac{1}{p_1p_2^2}+\dfrac{1}{p_2}+\dfrac{1}{p_2^2}=2\\ \dfrac{p_1p_2^2+p_1p_2+p_1+p_2^2+p_2+1}{p_1p_2^2}=2\\ p_1p_2^2-p_1p_2-p_1=p_2^2+p_2+1\\ p_1(p_2^2-p_2-1)=p_2^2+p_2+1 \end{eqnarray*}

Therefore, $p_1\equiv -1\bmod{p_2}$. Now the only way that that is possible is when $p_2=2$. Solving for $p_1$, we get that $p_1=7$. Checking, the sum of the reciprocals of the divisors of $\boxed{28}$ indeed sum to 2, and 28 does have 6 factors.

Furthermore, if we rearrange $\textbf{(2)}$, we can see that the number would be the sum of all its divisors other than itself, making it a perfect number. Checking the perfect numbers, we find that the second perfect number, $28$, fits the required form and is indeed the solution.

See also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 12 		Followed by:
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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