Sam2304! superb example given by you... I simply loved that. You gave an example and that is the only way to give an answer as x is an integer greater than 1...

Thank You once again...

shantanu86 wrote:

x = a^3
taking 4th power of both sides..
x^4 = a^12 ... (1.)

x= b^4
taking 3rd power of both sides..
x^3 = b^12 ... (2.)

Now, dividing (1.) by (2.) we have-
x = (a/b)^12 ..

Here, a/b has to be an integer hence [C] is the correct choice.

Shantanu, I love this explanation. You start with the statements and works toward the stem.

While some of the other explanations offer still a valuable intuitive approach, there is a downside--they start with the stem (assuming its plausibility) and work toward the statements. Though this works in this example, it can be dangerous in general.

To expound, if we start with, for instance, 2^1, work up to 2^6 = 64, and then square that, the result is indeed an example of an integer raised to the 12 power. On it's face, this seems to satisfy the sufficiency requirement.

(2^6)^2= 2^12=4096.

This proves the existence of at least one integer equal a 3rd, 4th, and 12th power; however, the question isn't asking whether its possible for this to occur, it's asking whether the specific number x does and whether we can prove that based on the statements. Therefore, our logic must start at the statements and work toward the stem.

In a sense, the above approach starts with the answer a^12 and works backward to prove that a^12=b^4=c^3. In this case, we luck out--all a^12s equal b^4s and c^3s. But using this approach as your standard for DS will sometimes lead you astray as you're more likely to miss opportunities to prove a statement insufficient. This is especially true with inequalities.

Alternatively, Shantanu starts with the statements and works toward the stem. This way he proves, not just the possibility, but the rule.

Artfully done.

So just to recap--

1) x=(a*a*a) --> x^4 = (a*a*a)(a*a*a)(a*a*a)(a*a*a)
2) x=(b*b*b*b) --> x^3 = (b*b*b*b)(b*b*b*b)(b*b*b*b)

Both Together) (x^4)/(x^3) = (a^12)/(b^12)

Here's the kicker:
Since(x^4)/(x^3) = x^(4-3) = x^1
x =(a^12)/(b^12) = (a/b)^12

So x=(a/b)^12 where a is divisible by b, and therefore an integer. In a sense, this becomes a divisibility problem.

--mj.

1.x=m^3......NS

2.X=m^4.....NS

Both:
Substitute m^4 from st2 in st1.......x=12

substitute m^3 from st1 in st2.......x=12

Therefore x=m^12

There is a very easy explanation why both statements together are sufficient.

(1) x is a cube of integer, means that the prime factorization of x contains primes raised to powers that are multiples of 3 like 0, 3, 6, 9, etc.

(2) x is a 4th power of an integer, means that the prime factorization of x contains primes raised to powers that are multiples of 4

(1) and (2) taken together imply that the prime factorization of x contains primes at powers that are BOTH mulitples of 4 and multiples of 3, which means the powers are multiples of 12 (since 3 and 4 are not divisible by each other).
If the prime factorization of a number contains primes to powers multiples of 12, it means x is the 12th power of an integer.

I posted an explanation here:

https://www.beatthegmat.com/interget-pow ... 14665.html