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What is the remainder when the positive integer n is divided

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What is the remainder when the positive integer n is divided [#permalink] New post 17 Nov 2009, 09:50
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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
[Reveal] Spoiler: OA

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Re: Division by K - Remainder - Gmat Prep [#permalink] New post 17 Nov 2009, 10:44
pierrealexandre77 wrote:
Hi guys,

I'm really stuck with this question. Could you please help me to find the solution?


easiest way is just start substituting numbers for K

if K = 2 n = 27 27/2 leaves remainder 1
if k = 3 n = 64/3 leaves remainder 1
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Re: Division by K - Remainder - Gmat Prep [#permalink] New post 17 Nov 2009, 10:46
pierrealexandre77 wrote:
Hi guys,

I'm really stuck with this question. Could you please help me to find the solution?


given that both n and k are +ve integers and k>1 with option 1 for any value of k>1 the remainder will always be 1. hence suff

option 2 says nothing abt n.hence insuff

hence A
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Re: Division by K - Remainder - Gmat Prep [#permalink] New post 17 Nov 2009, 10:52
A: Sufficient as we get a unique value of remainder = 1 when for K=2,3,4,5.....
B: Not Sufficient.

Ans: A
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Re: Division by K - Remainder - Gmat Prep [#permalink] New post 17 Nov 2009, 10:53
kp1811 wrote:
pierrealexandre77 wrote:
Hi guys,

I'm really stuck with this question. Could you please help me to find the solution?


given that both n and k are +ve integers and k>1 with option 1 for any value of k>1 the remainder will always be 1. hence suff

option 2 says nothing abt n.hence insuff

hence A


I understand with the picking number methodology, but is there a formal way to prove that?
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Re: Division by K - Remainder - Gmat Prep [#permalink] New post 17 Nov 2009, 11:20
I tried to expand S1 as

\(K^3+3K+3K^2+1\) using formula for \((a+b)^3 = a^3+b^3+3a^2b+3ab^2\)
when we divide \(K^3+3K+3K^2+1\) by K

k^3 , 3K and 3k^2 can be evenly divided by K , the remaining 1 would be remainder since K >1; 1 divide K, we get remainder 1 for any k>1

answer A
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Re: Division by K - Remainder - Gmat Prep [#permalink] New post 17 Nov 2009, 19:00
pierrealexandre77 wrote:
kp1811 wrote:
pierrealexandre77 wrote:
Hi guys,

I'm really stuck with this question. Could you please help me to find the solution?


given that both n and k are +ve integers and k>1 with option 1 for any value of k>1 the remainder will always be 1. hence suff

option 2 says nothing abt n.hence insuff

hence A


I understand with the picking number methodology, but is there a formal way to prove that?


we can expand (k+1)^3 and see that all terms are divisible by k except 1 (where k>1) which is the remainder
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Re: Division by K - Remainder - Gmat Prep [#permalink] New post 20 Dec 2010, 02:07
we do not need to remember the formular. Do like this

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

all the element can be divided by K, 1 is left. Answer A
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Re: Division by K - Remainder - Gmat Prep [#permalink] New post 20 Dec 2010, 02:21
Expert's post
thangvietnam wrote:
we do not need to remember the formular. Do like this

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

all the element can be divided by K, 1 is left. Answer A


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

Given: \(n=qk+r\). Question: \(r=?\)

(1) \(n=(k+1)^3\) --> you don't actually need to expand at all, since you understand what will happen after expansion: all term but the last will have \(k\) as the multiple and thus will be divisible by \(k\) and the last term will be 1 and 1 divide by \(k\) yields remainder of 1 (as \(k>1\)). Sufficient.

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

Answer: A.
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Re: Division by K - Remainder - Gmat Prep   [#permalink] 20 Dec 2010, 02:21
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