* Remainder

[mazhar.hussain]

When k^4 is divided by 32, the remainder is 0. Which of the following could be the remainder when k is divided by 32?

A.1
B.2
C.4
D.6
E.7

I am stuck with this one. Please help!

[Ben Ku]

Please cite the source (author) of this problem. We cannot reply unless a source is cited (and, if no source is cited, we will have to delete the post!). Thanks.

[mazhar.hussain]

Hi,

I saw this question at a forum named gmatclub. The link is as follows.
http://gmatclub.com/forum/remainder-que ... 82309.html

I was not aware of the mentioned source policy of yours, if my post is against the policy, kindly do let me know, so that I may delete it.

If the source is ok, do help me.

[snhassan1970]

I will go w/ 'C'.

K^4 is divisable by 32 --> both K^4 and K are even. --> the remainder can't be an odd number when K is divided by 32. We have to pick among 2, 4 and 6.

Also, As K^4 is divisable by 32, it is devisable by 2 and 4 (and 8 and 16). NOT 6. Lets take an example to pick the answer choices.

As K is even, lets assume k = 2, then k^4 = 2x2x2x2=16 --> Impossible. So, K can't be 2, the minimum vale of k is 4. Now lets take 4. then 4^4 = 256 divisable by 32 and lets take 6. --> 6^4 = 1296.--Not devisable by 32. The remainder is 16.

Any even number K takes it is divisable only when it is multiple of 4 (I have tried it in excel). Thus, the if K = 6, 10, 14 etc. Then K^4 is not divisable by 32. There is always some remainder and in every case the remainder is 16.

Thus the remainder could be multiple of 4 only.

Thus, 4 is the possible remainder. --> What is the official answer?

[mazhar.hussain]

As per the link shown above, the answer is "C".

But I am unable to pick your explanation.

[Ben Ku]

What did you do and where did you get stuck? What part of the explanations offered were you confused with? I think that will help direct a more helpful response.

Here's what I see:
If k^4 is divisible by 32, that means k^4 is divisible by 2 five times. So each of the k's must be divisible by 2; that takes care of four of the 2's. What about that last factor of 2?

That means one of the k's must also be divisible by 2 again, in other words, it's divisible by 4. If one of the k's is divisible by 4, that means all of them are divisible by 4.

When the problem states that k^4 is divisible by 32, you might as well say k^4 is divisible by 4^4 or 256, since k is divisible by 4.

If k is divisible by 4, then when it's divided by 32, it leaves a remainder that is also divisible by 4. The only possibility is (C). Hope that helps.