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Gud Explanation Fluke. Hope my explanations above were also correct although a little traditional

Oh, absolutely!! In fact, Kudos for that.

I believe you used the concept of mathematical induction, in which all terms but one are divisible by the denominator. I remember Karishma's describing it once. I don't remember that exactly.
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Gud Explanation Fluke. Hope my explanations above were also correct although a little traditional

Oh, absolutely!! In fact, Kudos for that.

I believe you used the concept of mathematical induction, in which all terms but one are divisible by the denominator. I remember Karishma's describing it once. I don't remember that exactly.

Thanks for my first kudos buddy. 'Mathematical Induction' Wats dat? Never Heard of that??

Thanks for my first kudos buddy. 'Mathematical Induction' Wats dat? Never Heard of that??

Responding to a PM:

Actually it was a discussion on 'Binomial Theorem' (Induction is an altogether different concept which is out of GMAT scope) Binomial theorem comes in handy in many remainder questions.

With a power of 3, it is easy to expand the expression and see that only 1 will be the remainder (as GMATPASSION did). For higher powers, binomial theorem can be used. I have put up a post on the Veritas blog discussing it and its applications. Here is the link. Get back in case there are any doubts.

Thanks for my first kudos buddy. 'Mathematical Induction' Wats dat? Never Heard of that??

Responding to a PM:

Actually it was a discussion on 'Binomial Theorem' (Induction is an altogether different concept which is out of GMAT scope) Binomial theorem comes in handy in many remainder questions.

With a power of 3, it is easy to expand the expression and see that only 1 will be the remainder (as GMATPASSION did). For higher powers, binomial theorem can be used. I have put up a post on the Veritas blog discussing it and its applications. Here is the link. Get back in case there are any doubts.

What is the remainder when the positive integer n is divided by the positive integer k, where k>1

(1) \(n=(k+1)^3= k^3 + 3k^2 + 3k + 1=k(k^2+3k+3)+1\) --> first term, \(k(k^2+3k+3)\), is obviously divisible by \(k\) and 1 divide by \(k\) yields the remainder of 1 (as \(k>1\)). Sufficient.

(2) \(k=5\). Know nothing about \(n\), hence insufficient.

The expansion of \((K+1)^3\) needs to be memorized? BTW great explanation @fluke

In case you do forget the expansion/don't know it, just multiply:

\((K+1)^3 = (K+1)(K^2 + 2K + 1)\) (We certainly know the expansion of \((K+1)^2\) or we can find it my multiplying (K+1)(K+1)) \((K+1)^3 = K^3 + 3K^2 + 3K + 1\)
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