The product of four different positive integers is 360 . What is the maximum possible sum of these four integers?
(A) 68
(B) 66
(C) 52
(D) 39
(E) 24
For the sum to be a maximum, we must choose the three smallest divisors in an effort to make the fourth divisor as large as possible.

The smallest 3 divisors of 360 are 1, 2 and 3, making $\frac{360}{1 \times 2 \times 3}=60$ the fourth divisor.

We note here that 1, 2 and 3 are the smallest three different divisors of 360 .

Therefore, it is not possible to use a divisor greater than 60, since there aren't three divisors smaller than 1,2 and 3.

Replacing the divisor 60 with a smaller divisor will decrease the sum of the four divisors. To see this, we recognize that the product of 3 different positive integers is always greater than
or equal to the sum of the 3 integers. For example, $1 \times 2 \times 4=8>1+2+4=7$.

The next largest divisor less than 60 is 45 , thus the remaining three divisors would have a product of $360 \div 45=8$, and therefore have a sum that is less than or equal to 8 .

This gives a combined sum that is less than or equal to $45+8=53$, much less than the previous sum of $1+2+3+60=66$. In the same way, we obtain sums smaller than 66 if we consider the other divisors of 360 as the largest of the four integers. Therefore, the maximum possible sum is 66 .

ANSWER: (B)