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by mathbyvemuri » Sun May 13, 2012 3:17 am
n = p^2q
n is a multiple of 5 and p is a prime => p must be 5
=>p^2 is 25
IMO "A"
But option "D" is also a possible one as p^2q^2 is a factor of p^2, ie.. 25
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by GMATGuruNY » Sun May 13, 2012 3:23 am
fangtray wrote:If n is a multiple of 5 and n = p^2q, where p and q are prime numbers, which of the following must be a multiple of 25?
a. p^2
b. q^2
c. pq
d. P^2q^2
e. p^3q
When a question asks for WHAT MUST BE X, try to prove that four of the answer choices DO NOT HAVE TO BE X.
The correct answer will be the remaining answer choice.

In order for n to be a multiple of 5, either p and/or q must be a multiple of 5.
Since the goal is to prove that four of the answer choices do NOT have to be a multiple of 25, start with the SMALLEST POSSIBLE COMBINATIONS.

Case 1: Let p=2 and q=5, so that n = 2²(5) = 20.
A) p² = 2² = 4. Not a multiple of 25. Eliminate A.
B) q² = 5² = 25. 25 is a multiple of 25. Hold onto B.
C) pq = 2*5 = 10. Not a multiple of 25. Eliminate C.
D) p²q² = 2²(5²) = 100. 25 is a multiple of 25. Hold onto D.
E) p³q = (2³)5 = 40. Not a multiple of 25. Eliminate E.

Case 2: Let p=5 and q=2, so that n = (5²)2 = 50.
B) q² = 2² = 4. Not a multiple of 25. Eliminate B.

The correct answer is D.
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by fangtray » Sun May 13, 2012 2:46 pm
GMATGuruNY wrote:
fangtray wrote:If n is a multiple of 5 and n = p^2q, where p and q are prime numbers, which of the following must be a multiple of 25?
a. p^2
b. q^2
c. pq
d. P^2q^2
e. p^3q
When a question asks for WHAT MUST BE X, try to prove that four of the answer choices DO NOT HAVE TO BE X.
The correct answer will be the remaining answer choice.

In order for n to be a multiple of 5, either p and/or q must be a multiple of 5.
Since the goal is to prove that four of the answer choices do NOT have to be a multiple of 25, start with the SMALLEST POSSIBLE COMBINATIONS.

Case 1: Let p=2 and q=5, so that n = 2²(5) = 20.
A) p² = 2² = 4. Not a multiple of 25. Eliminate A.
B) q² = 5² = 25. 25 is a multiple of 25. Hold onto B.
C) pq = 2*5 = 10. Not a multiple of 25. Eliminate C.
D) p²q² = 2²(5²) = 100. 25 is a multiple of 25. Hold onto D.
E) p³q = (2³)5 = 40. Not a multiple of 25. Eliminate E.

Case 2: Let p=5 and q=2, so that n = (5²)2 = 50.
B) q² = 2² = 4. Not a multiple of 25. Eliminate B.

The correct answer is D.
thanks ! this clears it up. can we derive anything from the fact that 5*x = p*p*q?
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by mathbyvemuri » Sun May 13, 2012 8:51 pm
I thought p^2q as p^(2q). That's why there was ambiguity in my explanation. It's better to represent it as qp^2.
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by Tdot23 » Wed May 23, 2012 4:41 am
To mathbyvemuri,
Its my bad for not better writing out the question - I did write this super quickly before my coworkers got into work. Thakns for posting the answer thought!
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by Anurag@Gurome » Wed May 23, 2012 4:44 am
fangtray wrote:If n is a multiple of 5 and n = p^2q, where p and q are prime numbers, which of the following must be a multiple of 25?
a. p^2
b. q^2
c. pq
d. P^2q^2
e. p^3q

If n is multiple of 5, and n = p²q where p and q are prime, then either p or q or both of them must be equal to 5. Let's analyze each of the cases. (Note that only one of the following can happen at a time)
1. p = 5, p² is multiple of 25, q² not
2. q = 5, q² is multiple of 25, p² not
3. p = q = 5, p² = q² = multiple of 25

We have to find a generalized expression containing p and q such that it becomes multiple of 25. From above analysis we know p² or q² is not that expression as they may or may not be a multiple of 25. But in p²q² both of them are present and simultaneously all the three cases are merged into one! For any of the above cases p²q² will be always a multiple of 25.

The correct answer is D.
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