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seofah
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seofah
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gameCode
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That is strange, I had thought it to be C as well. But there is one thing i have noticed.
When k=2 remainder =1,
k=3 r=0
for k =>4 , r=3

But what difference does it make ! The QA c makes sense, else i has to be E.
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seofah
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There was a a small typo in the question :?
I got the answer that is in line with OA, but any attempts are welcome. :)
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FN
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I get A..

What is the remainder when the positive integer n is divided by the positive integer k,
where k > 1?
(1) n = (k+1)^3
(2) k = 5


2) insuff..

1) n= (K+1)(K+1)^2= (K+1)(K^2+2k+1)= K^3+3K^2+K+3K+1

as you can N= will always be 1 greater than K..

therefore 1 is the remainder...

1 is suff
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A. You can pick k=2, k=3, k=4 and you'll see that the remainder is always 1.
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x2suresh
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fresinha12
I get A..

What is the remainder when the positive integer n is divided by the positive integer k,
where k > 1?
(1) n = (k+1)^3
(2) k = 5


2) insuff..

1) n= (K+1)(K+1)^2= (K+1)(K^2+2k+1)= K^3+3K^2+K+3K+1

as you can N= will always be 1 greater than K..

therefore 1 is the remainder...

1 is suff

agree with this solution
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botirvoy
What is the remainder when the positive integer n is divided by the positive integer k,
where k > 1?
(1) n = (k+1)^3
(2) k = 5


k=2,5 can be used to test. The remainder is always 1.
Rigorous proof is as follows:
n=k^3+3k^2+3k+1
Remainder is clearly 1.
1 is sufficient and hence A.



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