Is the positive integer n divisible by 6?

Hey King,

I was having trouble with this question and didn't understand where you got "a" from. Also, how did you know "n=1 when n is not divisible by 6" ?

Thanks

Official Solution:

We must determine whether \(n\) is evenly divisible by \(6\). In other words, we must determine whether \(n\) has a factor of 2 and a factor of 3 (since \(2 \times 3 = 6\)). If \(n\) is evenly divisible by 6, then \(\frac{n}{6} = k\) for some integer \(k\), and we can rewrite \(n\) as \(6k\).

Statement 1 says that \(\frac{n^2}{180}\) is an integer. In order for \(n\) to be divisible by 6, \(n^2\) must have two factors of 6, since \((6k)^2 = 36k^2\). Since \(n^2\) is divisible by 180, 180 is a factor of \(n^2\), and all factors of 180 are also factors of \(n^2\). Because 180 has 36 as a factor, and \(36 = 6 \times 6\), \(n^2\) has two factors of 6. Thus, \(n\) has 6 as a factor, and \(n\) is divisible by 6. Statement 1 is sufficient to answer the question. Eliminate answer choices B, C, and E. The correct answer choice is either A or D.

Statement 2 says that \(\frac{144}{n^2}\) is an integer. The prime factorization of 144 is \(2 \times 2 \times 2 \times 2 \times 3 \times 3\). This means that \(n^2\) can contain up to four factors of 2 and two factors of 3 (note that any given factor must occur an even number of times in the prime factorization of \(n^2\)). If \(n = 4\), the statement is satisfied, since \(\frac{144}{n^2} = \frac{144}{4^2} = \frac{144}{16} = 9\). In this case, \(n\) is NOT divisible by 6. However, \(n\) could also equal 6, since \(\frac{144}{n^2} = \frac{144}{6^2} = \frac{144}{36} = 4\). In this case, \(n\) is divisible by 6. Statement 2 does NOT provide sufficient information to answer the question.


Answer: A.

Hello DangerPenguin,

If n^2/180 is an integer, you can say n^2 = 180 * a [Where is a can be an integer e.g: 1,2,3 ... ]

The minimum value could be 180 * 1 [ where a=1 ] = 36 * 5 = 6^2 * 5

But 180 , which is 6^2 * 5 is not a perfect square. Therefore, the minimum value of n^2 should be = 180 * 5 = 6^2*5^2

Therefore n = 6 * 5 . Now this number n is divisible by 6. Isn't it ?


Coming to your next question, n = 1 and n =12 are the two values I picked to ensure that we are not getting definitive "Yes" or "No" by using the statement 2.

Let me know if you still have doubts

Thanks

To check => n is divisible by 6 => it must be divisible y 2 and 3
Statement 1 => Using the property that Both n and n^x have the same prime factors.=> n must have 2,3,5 as its prime factors
Although it may have more prime factors too but these are a must .
Hence n is clearly divisible by 6 => suff
Statement 2 => n=1 => NO n is not divisible by 6
N=6 =< YES n is divisible by 6
Contradiction
Insuff
Smash that A