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Re: If k is a multiple of 24 but not a multiple of 16, which of [#permalink]
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sandy wrote:
If k is a multiple of 24 but not a multiple of 16, which of the following cannot be an integer?

(A) \(\frac{k}{8}\)
(B) \(\frac{k}{9}\)
(C) \(\frac{k}{32}\)
(D) \(\frac{k}{36}\)
(E) \(\frac{k}{81}\)



If K is not a multiple of 16, then K won't be multiple of any 16x
You can test some numbers, but if you have no time, just scan the answer choices and see that only option C has a denominator 2*16, thus dividing K by 32 won't give us an integer
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Re: If k is a multiple of 24 but not a multiple of 16, which of [#permalink]
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Excellent question.
This is best solved using factors and logic:

We are told that k is some multiple of 24, but it is not a multiple of 16.
Implying:
k=24z
k=8*3*z
Now, as we are told that it is never a multiple of 16, it will never have 4 twos. By virtue of being a multiple of 24, k already has 3 twos - but it can never get another two.

Looking at the answer choices:

A) 8 is a factor of 24, so it is always an integer
B) This is possible, should k have another 3
C) Not possible. For k to be divisible by 32, it needs 5 twos. However, we know that it cannot get 4 twos - so there will never be enough twos to fully divide K.
D) Possible. 36=6^2=2^2*3^2, and k can certainly have enough 3s to divide into 36.
E) Possible. 81=3^4, k just needs enough 3s.

C
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Re: If k is a multiple of 24 but not a multiple of 16, which of [#permalink]
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Re: If k is a multiple of 24 but not a multiple of 16, which of [#permalink]
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