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np - prime numbers

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by Matt@VeritasPrep » Tue Jun 23, 2015 6:03 pm
nikhilgmat31 wrote:I think question should explicity tell that 4 distinct factors.

lets p=3
3^3 leads 3*3*3 which can have factors as - 1,3,3,3,9,27 which is not 4 which leads to Answer C.

What's say of others.
On the GMAT, assume that "number of factors" = "number of distinct factors".
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by nikhilgmat31 » Tue Jun 23, 2015 9:14 pm
seems I am confused.
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by deepak4mba » Mon Apr 02, 2018 11:17 pm
E
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by Jeff@TargetTestPrep » Thu Apr 05, 2018 4:19 pm
sairamGmat wrote:If n and p are different positive prime numbers, which of the integers n^4, p^3 and np has (have) exactly 4 positive divisors?
(A) n^4 only
(B) p^3 only
(C) np only
(D) n^4 and np
(E) p^3 and np
We may recall that when we add one to the exponent of each unique prime of a number, and then multiply together the sums, we get the total number of factors.

Thus, we see that n^4 has 4 + 1 = 5 factors and p^3 has 3 + 1 = 4 factors.

Since n x p = n^1 x p^1, so that product has (1 + 1)(1 + 1) = 4 factors.

Answer: E

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