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divisors of 64

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by j_shreyans » Wed Sep 17, 2014 1:50 am
If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?

A)1/3
B)1/2
C)5/7
D)10/21
E)15/28

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by GMATGuruNY » Wed Sep 17, 2014 2:42 am
j_shreyans wrote:If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?

A)1/3
B)1/2
C)5/7
D)10/21
E)15/28
P = (good options)/(all options).

Factors of 64:
1*64
2*32
4*16
8*8.
Total number of distinct factors = 7.

All:
From the 7 distinct factors above, the number of ways to choose 2 =7C2 = (7*6)/(2*1) = 21.

Good = distinct pairs with a sum less than 32:
1+2, 1+4, 1+8, 1+16
2+4, 2+8, 2+16
4+8, 4+16
8+16
Good options = 10.

Result:
P(pair with a sum less than 32) = (good options)/(all options) = 10/21.

The correct answer is D.
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