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# What is the reminder when the positive integer n is divided

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What is the reminder when the positive integer n is divided [#permalink]

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13 Oct 2008, 10:06
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What is the reminder when the positive integer n is divided by the postivie integer k, where k>1?

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

(2) k=5

OA=A

Thank you!

[Edited: the OA is A and not C]
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Last edited by JohnLewis1980 on 13 Oct 2008, 10:50, edited 1 time in total.

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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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13 Oct 2008, 10:40
For statement 1, if K > 1, k could be 2, 3, 4, 5, 6, 7, etc. With k being any of those numbers, when you put K, in the formula and you find n, you can have different remainders for n/k. Therefore not sufficient.

For statement 2, k is 5 we can't find n. Therefore not sufficient.

For both statements, if k is 5, you will get one answer for n. Therefore you can find the remainder for n/k.

Hope that helps.

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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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13 Oct 2008, 10:49
Sorry Piper, I put the wrong OA.

The right one is A and not C.

I edit the original post

(C is what I chose)

Cheers
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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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13 Oct 2008, 10:55
JohnLewis1980 wrote:
What is the reminder when the positive integer n is divided by the postivie integer k, where k>1?

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

(2) k=5

OA=A

Thank you!

[Edited: the OA is A and not C]

I dont think OA can be A..

n=(K+1)^3, suppose K=2 then 3^3=27/3 remainder is 0..but if K=4, then 5^3/3 has a non-zero remainder..

I think C is the correct answer..

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Senior Manager
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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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13 Oct 2008, 11:03
mmm, it could be possible that the GMAT PrepSoft were wrong?

I attached an screen with the OA

Cheers
Attachments

Reminder answer.JPG [ 43.73 KiB | Viewed 1397 times ]

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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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13 Oct 2008, 11:04
nikhilpoddar wrote:
A

Thank you
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Current Student
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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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13 Oct 2008, 11:14
JohnLewis1980 wrote:
mmm, it could be possible that the GMAT PrepSoft were wrong?

I attached an screen with the OA

Cheers

OOPs...OA is Correct..

n=(K+1)^3, suppose n=3^3 then k=2 then remainder is 1..

you will always note that remainder is 1..

A is correct

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Senior Manager
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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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13 Oct 2008, 12:12
yes, now it's clearer

This kind of problem is easier to solve using a couple of examples

Cheers
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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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13 Oct 2008, 12:16
From stmt1: n = k^2 + 2k + 1 and since the first two terms are divisible by k, the remainder will be 1. Hence, stmt 1 is sufficient.

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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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14 Oct 2008, 08:50
hi,
n=(k+1)3 and not (k+1)2 to break it into k^2 +2k+1.
however, (k+1)3 is k^3+1+3k^2+3k--[a^3 +b^3+3ab(a+b)]

the remainder will always be 1.

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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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14 Oct 2008, 10:50
scthakur wrote:
From stmt1: n = k^2 + 2k + 1 and since the first two terms are divisible by k, the remainder will be 1. Hence, stmt 1 is sufficient.

scthakur...how did you arrive at n = k^2 + 2k + 1 from n = (k+1)^3 ?

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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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15 Oct 2008, 01:12
Answer is A. The remanider is always 1. Reason being that (k+1)^3 = K^3+3k^2+2k+1. Hence k is common in all variable except 1. Hence 1 will always be the remainder.

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Re: DS: Reminder (source GMAT Prep Soft) [#permalink]

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15 Oct 2008, 01:29
bigtreezl wrote:
scthakur wrote:
From stmt1: n = k^2 + 2k + 1 and since the first two terms are divisible by k, the remainder will be 1. Hence, stmt 1 is sufficient.

scthakur...how did you arrive at n = k^2 + 2k + 1 from n = (k+1)^3 ?

My fault. I wrongly read as (k+1)^2. But, even with k+1)^3, all the terms but the last will have k and hence the remainder will always be 1.

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Re: DS: Reminder (source GMAT Prep Soft)   [#permalink] 15 Oct 2008, 01:29
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