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# GRE Math Challenge #3- Solve (s^3)(t^3) = v^2

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GRE Math Challenge #3- Solve (s^3)(t^3) = v^2 [#permalink]  06 Aug 2014, 14:03
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Question Stats:

32% (00:32) correct 67% (01:11) wrong based on 28 sessions
Level 170 question

Given $$(s^3)(t^3) = v^2$$ .If s and t are both primes, how many positive divisors of v are greater than 1, if v is an integer?

(A) two

(B) three

(C) five

(D) six

(E) eight
[Reveal] Spoiler: OA

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Re: GRE Math Challenge #3 [#permalink]  06 Aug 2014, 23:58
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s,t = 2 and v = 8. Positive divisors greater than 1 = {2,4,8}.
Therefore, answer would be three (B)
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Re: GRE Math Challenge #3- Solve (s^3)(t^3) = v^2 [#permalink]  17 Jul 2018, 15:04
soumya1989 wrote:
Level 170 question

Given $$(s^3)(t^3) = v^2$$ .If s and t are both primes, how many positive divisors of v are greater than 1, if v is an integer?

(A) two

(B) three

(C) five

(D) six

(E) eight

Tough one. Nice question. Tricky.

Note :

if we add 1 to the exponents of each primes and multiply them we get the total number of factors including 1 and the number itself.

Total factor : (3+1) (3+1) = 16.

Remember that 16 is the factor of $$v^2$$. So, v has 4 factors for sure as v's factors was squared.

So, v has 4 factors.

Now , it's stated in the question that we need to find out the number of divisors of v that are greater than 1. In our factor calculation we consider 1 too. So it's time to deduct now.

Total number of factors greater than 1 = 4-1 = 3.

Re: GRE Math Challenge #3- Solve (s^3)(t^3) = v^2   [#permalink] 17 Jul 2018, 15:04
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