A number N is divisible by both p and q, where p and q are integers gr

Question

A number N is divisible by both p and q, where p and q are integers greater than 1. Which of the following need not be a factor of N^3?

A. p^2
B. q^3
C. pq
D. (p^2)q
E. (p^2)(q^2)

gmatophobia · Answer

This is a 'can be true' type of question.

Let's assume  p = 2 ,  q = 4  and  N = 4

N^3 = 4 * 4 * 4 = 2^6

From the given options we can see that,  N^3  may be divisible by all except Option E.

Option E

lomachinsky01 · Answer

Why (p^2)(q^2) does not need to be a factor of N^3?

If p=2 and q=4 then (p^2)*(q^2) = 2^2 * 4^2 = 4^3 thar is the exact value of N^3. So N^3 could be divisible by (p^2)*(q^2).

Take for instance this scenario: p=2 and q=3. In this case, N=6 ==> N^3=216. Factoring 216 we get 2^3 * 3^3 and (p^2)(q^2) could be a factor of N^3.

g_giaimo02 · Answer

We have N*N*N
and we know that for sure p is a factor of N and q is a factor of N, but we can't know if pq is a factor of N as well. If u think about the answers:
p^2 will always be a factor of N^3 (we'll have p*p*p as factors)
q^3 same reasoning (q*q*q as factors)
pq same
p^2*q will always be a factor because we can take 2 times p from N*N and one time q from the last N

finally 
p^2*q^2 would be p*p*q*q, since we only have N*N*N we can say that this doesn't need to be a factor. (It will be a factor only if pq is a factor of N)

Samya2113 · Answer

KarishmaB   Bunuel  Can you please help with solving this question?

Thanks!

KarishmaB · Answer

N is divisible by both p and q.

If p and q are coprime, then N is of the form pqk. Say p = 2, q = 3, then N = 6k. In this case all given options will be factors of N^3.

But if p and q have a common factor, then N needn't be of the form pqk.
Say p = 2, q = 6 and N = 18. Both 2 and 6 are factors of 18 but they have a 2 common and N has only one 2 so pq (= 12) is not a factor of N.

N^3  will have three of the common factor 2. So p^2, q^3, pq, (p^2)q and p(q^2) will be factors of N^3 but p^2*q^2 may not be.

N^3 = 18^3 = 2^3 * 3^6 
 p^2q = 2^2 * 2 * 3  (factor of N^3)
 p^2 q^2 = 2^2 * 2^2*3^2  (Has four 2s but N^3 has only three 2s. Not a factor of N^3)

Answer (E)

bumpbot · Answer

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A number N is divisible by both p and q, where p and q are integers gr

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A number N is divisible by both p and q, where p and q are integers gr

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