Hi swerve,

We're asked to find the number of factors of a 3-digit EVEN number XYZ where X, Y and Z are distinct prime numbers (2, 3 and 5). Since the number has to be EVEN, the last digit MUST be 2 (and the remaining 2 digits are either 35 or 53). You might find it help to TEST a VALUE and prime factor it into its pieces:

IF the number is 352, then the prime factorization would be:
(2)(176)
(2)(2)(88)
(2)(2)(2)(44)
(2)(2)(2)(2)(22)
(2)(2)(2)(2)(2)(11)

Now that we have the prime factorization, we can list all of the factors:
1 and 352
2 and 176
4 and 88
8 and 44
16 and 22
32 and 11

Total factors = 12

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

swerve wrote:Find the number of factors of a three digit even number xyz where x, y and z are distinct prime numbers (2, 3 and 5)

A. 8
B. 10
C. 12
D. 16
E. 24

The OA is C.

Please, can any expert explain this PS question for me? I tried to solve it but I need your help. Thanks.

Since xyz is even and x, y, and z are distinct prime numbers (2, 3, and 5), xyz can either be 352 or 532.

Recall that to determine the number of factors of a number, add 1 to each exponent of its prime factorization and then multiply those numbers.

If xyz = 352 = 4 * 88 = 2^2 * 2^3 * 11 = 2^5 * 11, then it has (5 + 1) * (1 + 1) = 6 * 2 = 12 factors.

If xyz = 532 = 4 * 133 = 2^2 * 7 * 13, then it has (2 + 1) * (1 + 1) * (1 + 1) = 3 * 2 * 2 = 12 factors.

Therefore, we can see that either way, xyz has 12 factors.

Answer: C