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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 25 Jun 2010, 22:42
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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

Last edited by Bunuel on 12 Nov 2012, 02:26, edited 1 time in total.
Renamed the topic and edited the question.
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Re: GMAT prep DS- Remainder [#permalink] New post 26 Jun 2010, 07:45
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What is the remainder when the positive integer n is divided by the positive integer k, where k>1

(1) \(n=(k+1)^3= k^3 + 3k^2 + 3k + 1=k(k^2+3k+3)+1\) --> 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.

Answer: A.
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Re: GMAT prep DS- Remainder [#permalink] New post 26 Jun 2010, 16:06
Yep A. Nice explanation Bunuel.
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Re: GMAT prep DS- Remainder [#permalink] New post 29 Jun 2010, 18:33
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bunuel's solution is certainly the quickest way to solve. (Of course, it would be (k+1)^3 = k^3 + 3k^2 + 3k + 1, and since each of the "k" terms is divisible by k, the remainder is 1).

But if this deduction/concept didn't jump out at you, you could also solve fairly quickly by picking numbers. If we pick different integer values for k and always end up with the same remainder for n/k, then we can trust that (1) is sufficient. On the other hand, if ever we get different remainder values, we know at once (1) is insufficient:

(1): let k=2. Then, n = (2+1)^3 = 27. 27/2 leaves a remainder of 1.
----let k=3. Then, n = (3+1)^3 = 64. 63 is clearly divisible by 3. Thus, 64/3 leaves a remainder of 1 again.
----let k=4. Then, n = (4+1)^3 = 125. 124 is clearly divisible by 4. Thus, 125/4 leaves a remainder of 1 again.

We've convinced ourselves that (1) is sufficient!
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Re: DS question [#permalink] New post 31 Jul 2010, 06:57
Samarth0711 wrote:
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

Answer given is (A) - Need solution.



1 sufficient

n=3k + 3 ==> reminder is 3
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Re: DS question [#permalink] New post 31 Jul 2010, 09:25
Its (k+1) cubed and not 3 multiplied by (K+1)
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Re: DS question [#permalink] New post 31 Jul 2010, 09:29
Samarth0711 wrote:
Its (k+1) cubed and not 3 multiplied by (K+1)


it is likely the same

n=(k+1)^3 ==> n=k^3 + 3k²+3k+1 ==> n=k(K²+3k+3) +1 so the reminder is 1

DO you agree ?
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What is the remainder when the positive integer n is divided [#permalink] New post 12 Nov 2012, 02:13
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
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Re: What is the remainder when the positive integer n is divided [#permalink] New post 12 Nov 2012, 10:03
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Re: What is the remainder when the positive integer n is divided [#permalink] New post 11 Nov 2013, 09:04
JoyLibs wrote:
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+1)^3

Take any value of k(for eg 8)
K= 8 ; 9^3/8 = (1^3)/9 = 1(because each value of 9/8 gives remainder 1.Hence 1 X 1 X 1)
k = 2 ; 3^3/2 = (1^3)/2 = 1
etc.A sufficient

b) Insufficient
OA A
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Re: What is the remainder when the positive integer n is divided [#permalink] New post 03 Jan 2015, 09:47
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Re: What is the remainder when the positive integer n is divided   [#permalink] 03 Jan 2015, 09:47
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