If two distinct positive divisors of 128 are randomly selected, what is the probability that their sum will be less than 64?	Choose the correct choice from the following
A) 2/5 B) 5/7 C) 9/14 D) 15/28 E) 17/35
The positive divisors are 1, 2, 4, 8, 16, 32, 64, 128.   If any of the first six are chosen, the sum will be less than 64.   P(sum is less than 64) = 6/8 * 5/7 = 15/28   The answer is 15/28.
<result>D</result>
If two distinct positive divisors of 128 are randomly selected, what is the probability that their sum will be less than 64?