2005 Alabama ARML TST Problems/Problem 13


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


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
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