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Bunuel
If a is a multiple of 3 and a = y*z^2, where y and z are prime numbers, which of the following must be a multiple of 9 ?

A. y^2
B. z^2
C. yz
D. y^2*z^2
E. y^3*z

We are given that a is a multiple of 3 and a = y(z^2), in which y and z are prime numbers. Thus, either y or z MUST be 3. We need to determine which of the answer choices MUST be true.

A) y^2

Answer choice A does not have to be true. For instance, if y = 2 and z = 3, then the given statement is satisfied and y^2 = 2^2 is not a multiple of 9.

B) z^2

Answer choice B does not have to be true. For instance, if y = 3 and z = 2, then the given statement is satisfied and z^2 = 2^2 is not a multiple of 9.

C) yz

Answer choice C does not have to be true. For instance, if y = 2 and z =3, then the given statement is satisfied and 2 x 3 = 6 is not a multiple of 9.

D) (y^2)(z^2)

Answer choice D will always be true. In order for a to be a multiple of 3 in the equation a = y(z^2), in which y and z are prime numbers, either y or z MUST be 3. If y = 3, then y^2 = 9, and hence regardless what z is, (y^2)(z^2) is a multiple of 9. If y ≠ 3, then z = 3. If z = 3, then z^2 = 9, and hence (y^2)(z^2) is a multiple of 9. Thus (y^2)(z^2) will ALWAYS be a multiple of 9. We can stop here.

Answer: D
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