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Archit3110
BrentGMATPrepNow
question is smallest +ve integer for which 288 is a factor of 6^n
6^n/2^3 *6^2

n has to be >2
least value of which would get us an integer value or say perfectly divide is 6^n/288 is when n = 5 and we get factor as 27
option D

Good catch!!
I was reading the question the opposite way around.
I read it as 6^n is a factor of 288 (not the other way around as intended)

I've fixed my solution accordingly.
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BrentGMATPrepNow
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What is the smallest positive integer \(n\) for which 288 is a factor of \(6^n\)?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
-------ASIDE---------------------------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving factors, we can say:

If k is a factor of N, then k is "hiding" within the prime factorization of N

Consider these examples:
3 is a factor of 24, because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a factor of 70 because 70 = (2)(5)(7)
And 8 is a factor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a factor of 630 because 630 = (2)(3)(3)(5)(7)
-----BACK TO THE QUESTION!---------------------

So, the original question can be rephrased to ask "What's the smallest value of n so that 288 is hiding in the prime factorization of 6^n?
Let's find out..

288 = (2)(2)(2)(2)(2)(3)(3)
So, in order for 288 to be a factor of 6^n, the prime factorization of 6^n must contain at least five 2's and two 3's

So, for example, 6⁴ = (6)(6)(6)(6) = (2)(3)(2)(3)(2)(3)(2)(3).
As you can see, 6⁴ contains at least two 3's, BUT it doesn't contain the five 2's we need.
So, 288 is not a factor of 6⁴

On the other hand, 6⁵ = (6)(6)(6)(6)(6) = (2)(3)(2)(3)(2)(3)(2)(3)(2)(3) = (2)(2)(2)(2)(2)(3)(3)(3)(3)(3), which means 288 is definitely a factor of 6⁵

Answer: D
Hi BrentGMATPrepNow, great solution, but I believe there's a small error in the translation of the problem at your end.

We need to the smallest value of '\(n\)' for which \(288 \) will be a factor of \(6^n\). The answer would be (D) \(5\) as \(288 = 2^5 * 3^2\). And \(6^n = (2*3)^n\). For \(288 \) to be a factor of \(6^n = (2*3)^n\), it should at least be equal to \(5\).

Option (D)
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Hi BrentGMATPrepNow, great solution, but I believe there's a small error in the translation of the problem at your end.

We need to the smallest value of '\(n\)' for which \(288 \) will be a factor of \(6^n\). The answer would be (D) \(5\) as \(288 = 2^5 * 3^2\). And \(6^n = (2*3)^n\). For \(288 \) to be a factor of \(6^n = (2*3)^n\), it should at least be equal to \(5\).

Option (D)
Thanks for the heads up, but I think you may be referring to my original solution (which I have since edited following @Archit3110's earlier comments)

Thanks nonetheless.
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288=2^5*3^2
For 6^n to be divided by 288
Soln:
6^n=2^n×3^n
The smallest value for n cannot be less than 5.
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