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Re: Number propierties [#permalink]
19 Sep 2011, 02:27

GMATPASSION wrote:

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. _________________

Re: Number propierties [#permalink]
19 Sep 2011, 02:30

fluke wrote:

GMATPASSION wrote:

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??

Re: Number propierties [#permalink]
19 Sep 2011, 21:22

2

This post received KUDOS

Expert's post

GMATPASSION wrote:

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.

Re: Number propierties [#permalink]
19 Sep 2011, 23:24

VeritasPrepKarishma wrote:

GMATPASSION wrote:

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.

Re: What is the remainder when the positive integer n is divided [#permalink]
04 Jul 2013, 02:12

1

This post received KUDOS

Expert's post

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.

Re: What is the remainder when the positive integer n is divided [#permalink]
04 Jul 2013, 19:07

1

This post received KUDOS

Expert's post

fozzzy wrote:

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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