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# (s^3)(t^3) = v^2 If s and t are both primes, how many positive diviso

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Joined: 02 Sep 2009
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(s^3)(t^3) = v^2 If s and t are both primes, how many positive diviso  [#permalink]

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01 Aug 2017, 00:23
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Question Stats:

59% (02:12) correct 41% (02:18) wrong based on 168 sessions

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(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

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Re: (s^3)(t^3) = v^2 If s and t are both primes, how many positive diviso  [#permalink]

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01 Aug 2017, 00:44
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Bunuel wrote:
(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

Let s & t be 2

2^6 = v^2
v = 2^3
v will have 4 divisors including 1
excluding 1 it will have 3

B
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Re: (s^3)(t^3) = v^2 If s and t are both primes, how many positive diviso  [#permalink]

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01 Aug 2017, 00:56
S and T will be equal for v to be an integer.
So V=S^3. And number of divisors will be 3+1=4
Since we are not counting 1 so answer will be B (three)

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Re: (s^3)(t^3) = v^2 If s and t are both primes, how many positive diviso  [#permalink]

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01 Aug 2017, 11:13
The easiest way in my opinion is to plug numbers and pick the right answer. Here is another way - more logical - to solve the question: As V^2 is a perfect square it has even number of each of his prime factors. As S and T are both prime numbers the multiplication of S^3 x T^3 has 3 S's and 3 T's. The only way for the aforementioned multiplication to has even numbers of primes is when S equal to T (3+3=6). Therefore we can write T^6=V^2 ----> T^3=V. As T is a prime number it has a total of 4 factors (3+1). We ask for the factors of V which are greater than 1, so we should exclude the case of T in a power of 0 (T=1). So the answer is 4-1=3.
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(s^3)(t^3) = v^2 If s and t are both primes, how many positive diviso  [#permalink]

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11 Jan 2018, 05:47
Given: v is an integer => $$\sqrt{s^3 * t^3}$$ => must be an integer

but given s and t are prime numbers, then $$\sqrt{s^3 * t^3}$$ cannot be an integer, unless and until both the primes are same.

for v to be an integer, s = t , $$v^2 = s^3 * s^3 = s^6$$
=> $$v = s^3$$
=> number of factors of v = (3 + 1) = 4
=> Number of factors of v greater than 1 = (4 - 1) = 3
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Re: (s^3)(t^3) = v^2 If s and t are both primes, how many positive diviso  [#permalink]

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18 Jan 2018, 07:41
1
Bunuel wrote:
(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

In order for the equation (s^3)(t^3) = v^2 to hold, we see that s and t must be equal;, thus, we can say:

(s^3)(s^3) = v^2

s^6 = v^2

s^3 = v

To determine the total number of factors of v, we can add 1 to the exponent of 3, and thus v has 3 + 1 = 4 total factors. Since one of those factors is “1”, v has 3 factors other than 1.

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Re: (s^3)(t^3) = v^2 If s and t are both primes, how many positive diviso  [#permalink]

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09 Feb 2019, 22:08
Bunuel wrote:
(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

I preferred plug in here,

(s^3)(t^3) = v^2 If s and t are both primes

s & t can be same => 2^6 = 8^2

8 has 3 divisors 2,4,8

Before marking the question, lets check another case

3^3 2^3 = 216 ! = a perfect square

5^3 * 2^3 = 1000 ! = a perfect square

3^6 = 9^2, this will again work

B
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Re: (s^3)(t^3) = v^2 If s and t are both primes, how many positive diviso   [#permalink] 09 Feb 2019, 22:08
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