Problem Solving

yvonne12 wrote:If N is a multiple of 5 on n=p^2q, where p and q are prime numbers, which must be a multiple of 25?

p^2
q^2
pq
p^2 q^2
p^3 q

A common phrase that is used on the GMAT is the word must. In this question, we are asked which of the following must be a multiple of 25. This means that one of our answer choices will always be a multiple of 25, no matter what. It is our job to determine which one, based on the given information.

We are given that n is a multiple of 5, n = (p^2)q, and that p and q are prime numbers.

Because n is a multiple of 5, a prime number, we know that either p or q is 5. Let's now analyze each answer choice to determine which one MUST (in all cases) be a multiple of 25.

A) p^2

If p = 3, then p^2 = 9 is not a multiple of 25. Answer choice A is not correct.

B) q^2

If q = 3, then q^2 = 9 is not a multiple of 25. Answer choice B is not correct.

C) pq

If p = 5 and q = 3 (or vice versa), pq = 15 is not a multiple of 25. Answer choice C is not correct.

D) (p^2)(q^2)

Regardless of which values we select for p and q, since we know that either p or q is 5, (p^2)(q^2) will always be a multiple of 25. If this is difficult to see, let's use numbers.

If p = 5 and q = 3, (p^2)(q^2) = (25)(9) is a multiple of 25.

If p = 3 and q = 5, (p^2)(q^2) = (9)(25) is also a multiple of 25.

Answer choice D is correct.

For practice, let's analyze answer choice E.

E) (p^3)q

If p = 3 and q = 5, then (p^3)q = 135 is not a multiple of 25. Answer choice E is not correct.

Answer: D

If n is multiple of 5, and n = p²q, where p and q are prime numbers, which of the following MUST be a multiple of 25?

A) p²
B) q²
C) pq
D) p²q²
E) p³q

-----ASIDE---------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k (i.e., N is a multiple of k), then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-----ONTO THE QUESTION!---------------------

If p and q are prime numbers, and p²q is divisible by 5, then either p = 5, q = 5 or they both equal 5.

We're looking for an expression that MUST be divisible by 25, which means there must be TWO 5's "hiding" in the prime factorization of the expression.

A) p²
We cannot be certain that there are TWO 5's "hiding" in this expression.
It could be the case that p = 2 and q = 5, in which case p² is NOT divisible by 25
ELIMINATE A

B) q²
We cannot be certain that there are TWO 5's "hiding" in this expression.
It could be the case that p = 5 and q = 2, in which case q² is NOT divisible by 25
ELIMINATE B

C) pq
We cannot be certain that there are TWO 5's "hiding" in this expression.
It could be the case that p = 5 and q = 2, in which case pq is NOT divisible by 25
ELIMINATE C

D) p²q²
YES, we can be certain that there are TWO 5's "hiding" in this expression.
If p = 5, then p²q² = 25q², which is DEFINITELY divisible by 25
If q = 5, then p²q² = 25p², which is DEFINITELY divisible by 25

E) p³q
We cannot be certain that there are TWO 5's "hiding" in this expression.
It could be the case that p = 2 and q = 5, in which case p³q is NOT divisible by 25
ELIMINATE E

Answer = D

Cheers,
Brent