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The only catch in this question is to remember that k is a positive integer such as 1 or 2 or 3 or so on...

Let's look at the statements:

Statement 1: If \(\sqrt{k}\) is divisible by 9, naturally k will be divisible by 3 also since \(3^2\)=9

Statement 2: Now, here's where the catch comes into play. If \(k^2\) is divisible by 3 and k is a positive integer, k will definitely be divisible by 3 too. Note here, that we cannot take k to be a value such as \(\sqrt{3}\), this is because k is a positive integer.

So, clearly, both statements are sufficient and answer is D
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Bunuel
Is a positive integer k is divisible by 3?


(1) \(\sqrt{k}\) is divisible by 9

(2) k^2 is divisible by 3

Statement 1:

We can write this as \(\sqrt{k} = 9*i\), where \(i\) is a nonnegative integer. Then we have \(k = 81*i^2\), hence k is a multiple of 81. Then k must be divisible by 3. Sufficient.

Statement 2:

We are given k is already an integer. If \(k^2\) is divisible by 3, the factor of 3 must come from a single k so k must have a factor of 3. Sufficient.
(Another way to think about this is we fill in the square, \(k^2\) must be divisible by 9 given it is divisible by 3).

Ans: D
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Bunuel
Is a positive integer k is divisible by 3?


(1) \(\sqrt{k}\) is divisible by 9

(2) k^2 is divisible by 3


M36-19

Official Solution:


Is a positive integer \(k\) is divisible by 3?

(1) \(\sqrt{k}\) is divisible by 9

\(\sqrt{k}=9x\), for a positive integer \(x\). Square: \(k = 81x^2= a \ multiple \ of \ 3\). Sufficient.

(2) \(k^2\) is divisible by 3

Exponentiation does not "produce" primes (meaning that if \(m\) and \(n\) are positive integers, then \(m^n\) will have the same prime factors as \(m\) itself). So, if 3 is a factor of \(k^2\), then 3 must also be a factor of \(k\). Sufficient.


Answer: D
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