Problem:
Let be the sum of the base logarithms of all of the proper divisors of . (By a proper divisor of a natural number we mean a positive integral divisor other than and the number itself.) What is the integer nearest to
Solution:
The number has distinct positive divisors. To see this, observe that they are all of the form ; thus there are seven choices for ( and, independently, the same seven choices for . Apart from , the other divisors form pairs such that the product of each pair is . Since one of these pairs consists of the improper divisors and , it follows that the product of all proper divisors of is or . Moreover, since the sum, , of the logarithms is equal to the logarithm of the product of these numbers, . The nearest integer to is clearly .
The problems on this page are the property of the MAA's American Mathematics Competitions