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X, Y, and Z are three different prime numbers, the product

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by BTGmoderatorLU » Sat Sep 07, 2019 12:14 am

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Source: GMAT Prep

X, Y, and Z are three different prime numbers, the product XYZ is divisible by how many different positive numbers?

A. 4
B. 6
C. 8
D. 9
E. 12

The OA is C
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by Brent@GMATPrepNow » Sat Sep 07, 2019 6:06 am
BTGmoderatorLU wrote:Source: GMAT Prep

X, Y, and Z are three different prime numbers, the product XYZ is divisible by how many different positive numbers?

A. 4
B. 6
C. 8
D. 9
E. 12

The OA is C
----ASIDE-------------------------
If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) =(5)(4)(2) = 40
-----------------------------------
Since X, Y and Z are different PRIME numbers, we can say: XYZ = (X^1)(Y^1)(Z^1)
So, the number of positive divisors of XYZ = (1+1)(1+1)(1+1) =(2)(2)(2) = 8

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Thu Sep 12, 2019 10:22 am
BTGmoderatorLU wrote:Source: GMAT Prep

X, Y, and Z are three different prime numbers, the product XYZ is divisible by how many different positive numbers?

A. 4
B. 6
C. 8
D. 9
E. 12

The OA is C
To determine the number of factors of XYZ, or (X^1)(Y^1)(Z^1), we add 1 to each exponent of each unique prime factor and then multiply those values together. The result will equal the number of factors of the given number.

Thus, XYZ has (1 + 1)(1 + 1)(1 + 1) = 2 x 2 x 2 = 8 factors.

Answer: C

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