M30-07

For what its worth, I had a very similar problem on the GMAT when I took it.

Here's my solution:

For something to be a perfect cube, all prime factors must have a power that is a multiple of 3.

Start by factoring 6,480
\(=80*81\)
\(=2^3*10*3^4\)
\(=2^4*3^4*5\)

Okay, so the smallest solution will move powers of 4 to 6 and the power of 1 to 3
I.E. we need \(2^2*3^2*5^2\)
= \(900\)

Thus,
\(\sqrt{n} = 900\)
\(\sqrt{n} = 30^2\)
\(n = 30^4\)

ANSWER E

I am not sure if this is still helpful but whenever we find the perfect cube of an integer we always need to ensure that the powers of the bases are multiples of 3.
Take for example 8 - cube root of 8 is 2 because 8 = 2*2*2. This function works in a similar manner to perfect square. Just like we get an integral square root only when the integer is a perfect square.
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Similarly, in this question - I need to find the value of \sqrt{n} such that the whole no => 6480*\sqrt{n} is a cube root.
Step 1 - break down each prime factor of 6480 (8*810). 810 = 10*9*9 (further breakdown 9 and 10) and 8 = 2*2*2.
Step 2 - once we find each prime factor we then add the exponents for the similar bases. (2^4*5^1*3^4).
Step 3 - now all we need to do is see which of the values of n in the answer options when taken \sqrt{n} will give the "extra" powers that is required by the whole expression to ensure that the powers of the bases are multiples of 3. For example - we need to add to 2^2 to the expression to make the power of the base 2 a multiple of 3. U can do a similar check for the other bases as well.

Hope this helps!

Bunuel I need a more in-depth explanation for this. Like where do you even get 2^2*3^2*5^2 and why is this relevant? And what do you mean by "complete the powers of 2,3, and 5 to equal a multiple of 3" and why does this matter? How does this end up getting the cube root? HELP!!

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

-Let \(M=6480*\sqrt{n}=k^3\)
-We can quickly eliminate A) and C) since these answers can't make M an integer.
-We know that the unit digit of M must be 0, so the unit digit of k must also be 0 to make M end with 0=>\(M=k^3\) must contain at least 3 zeros at the end of the number.
=>Only E) satisfies the condition.
The answer is E