If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?
Choose one.
A) 1/3  B) 1/2  C) 5/7  D) 10/21  E) 15/28
list the distinct positive factors: 1,2,4,8,16,32,64   selecting 2 from this list = 7C2 = 21   as per the given condition there sum has to be less than 32   picking any 2 numbers from our list of first five numbers will render the sum less than 32 in any case   selecting 2 number from 5 numbers = 5C2 = 10   proability = 10/21   correct answer - 10/21
<result>D</result>
If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?