2005 Alabama ARML TST Problems/Problem 13
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 , and let and be primes. Therefore, one of the following is true:
For the first one, the sum of the reciprocals of the divisors of is therefore . The smallest prime (2) makes that less than 2, and if gets bigger, then that expression gets smaller, so there is absolutely no way that . So the second case is true.
Therefore, . Now the only way that that is possible is when . Solving for , we get that . Checking, the sum of the reciprocals of the divisors of indeed sum to 2, and 28 does have 6 factors.
Furthermore, if we rearrange , 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, , 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 |