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Remainder prob

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Remainder prob

by ricaototti » Thu Aug 28, 2008 5:18 am
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: Start with just 1. If , we can expand this to , and now it’s clear that the remainder of n/k is 1. Thus the answer has to be A or D.

WHY? It is not clear to me why remainder must be 1. I really dont get it. Can someone please clarify it. Thanks
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Re: Remainder prob

by sudhir3127 » Thu Aug 28, 2008 5:26 am
ricaototti wrote:What is the remainder when the positive integer n is divided by the positive integer k, where k>1?
1)
2) k = 5

Answer: Start with just 1. If , we can expand this to , and now it’s clear that the remainder of n/k is 1. Thus the answer has to be A or D.
n = (k+1)^3
WHY? It is not clear to me why remainder must be 1. I really dont get it. Can someone please clarify it. Thanks
i go with D

u cant start with 1 as the question stem says k>1

thus u need to pick any number >1

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

pick k = 2

then n = 9

remainder is 1.....................suffcient

statmenent 2. k=5

n = 6^3

6^3/5 ~1.........hence sufficient

hope that helps..
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by ricaototti » Thu Aug 28, 2008 5:38 am
Ops.... you made the classical mistake of considering 1 when solving for 2. I guess answer is just A.

Can we solve 1 ( the first statement) without pulgging in?
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Got doubt!!!!!

by kshankker » Thu Aug 28, 2008 5:48 am
@ sudhir
statement 1 is okay i agree with u.
But statement 2 has problem....becoz it only specifies K=5...it only knows wat is k and moreover n is in 1st statement so u cant use in 2nd rt.....
if iam wrong pls help me........
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by sudhir3127 » Thu Aug 28, 2008 5:49 am
ricaototti wrote:Ops.... you made the classical mistake of considering 1 when solving for 2. I guess answer is just A.

Can we solve 1 ( the first statement) without pulgging in?
ur right !!!. sorrie abt that..

i think plugging in saves lot of time ...

yes answer shud be A
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Re: Remainder prob

by Ian Stewart » Thu Aug 28, 2008 7:57 am
ricaototti 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: Start with just 1. If , we can expand this to , and now it’s clear that the remainder of n/k is 1. Thus the answer has to be A or D.

WHY? It is not clear to me why remainder must be 1. I really dont get it. Can someone please clarify it. Thanks
The suggestion is to multiply (k+1)^3 out. If you do this, you should find:

n = (k+1)^3 = k^3 + 3k^2 + 3k + 1

Notice that k^3 + 3k^2 + 3k is a multiple of k. So n is 1 larger than a multiple of k, and the remainder will be 1 when n is divided by k.

Statement 2) is clearly insufficient on its own, so the answer should be A.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Re: Remainder prob

by aj5105 » Fri Jan 16, 2009 2:24 am
cool stuff Ian..
Ian Stewart wrote:
ricaototti 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: Start with just 1. If , we can expand this to , and now it’s clear that the remainder of n/k is 1. Thus the answer has to be A or D.

WHY? It is not clear to me why remainder must be 1. I really dont get it. Can someone please clarify it. Thanks
The suggestion is to multiply (k+1)^3 out. If you do this, you should find:

n = (k+1)^3 = k^3 + 3k^2 + 3k + 1

Notice that k^3 + 3k^2 + 3k is a multiple of k. So n is 1 larger than a multiple of k, and the remainder will be 1 when n is divided by k.

Statement 2) is clearly insufficient on its own, so the answer should be A.
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by gaggleofgirls » Fri Jan 16, 2009 10:26 am
For 1) I multiplied out the numerator (k+1) ^3 and got:
K^3 + 3K^2 + 3K + 1 / K

Since K^3, 3K^2 and 3K are all multiples of K, then they will all divide evenly by K leaving the 1 as the remainder.

So now the only choices are A and D.

2) by itself gives us only information on the denominator, so not sufficient.

hence the answer is A.

-Carrie
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by aanchalsinha » Sat Aug 20, 2011 1:40 am
I know this question was posted 3 years ago but I'd still appreciate a response.

I had a question regarding statement 1, plugging in isnt the right approach to this then. Because when I use k = 2, the remainder is 1 however if k = 3, the remainder is 0.

I didnt think of opening out the cube. Is there a way of knowing when to expand the values? or should u do that regardless?

Thanks
Aanchal
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by aanchalsinha » Sat Aug 20, 2011 1:45 am
As I type it out, i understood my mistake. sorry for the post
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by n2739178 » Tue Nov 08, 2011 1:35 pm
u can also use scenario charts for statement 1...

k (k + 1)^3 n n/k
----------------------
2 27 27 13 1/2
3 64 64 21 1/3 etc.
4
5
6

notice that n/k always results in a remainder of 1 ... e.g. 13 1/3, 21 1/3 etc.
Last edited by n2739178 on Fri Nov 11, 2011 2:40 am, edited 1 time in total.
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by IWillSurvive » Tue Nov 08, 2011 2:45 pm
I know this thread is somewhat old, but this is a great example of those tricky remainder problems. I was already prepared to say "B", but now I can see why it is D! You almost have to test numbers for these...just to check scenarios. One thing I did was check odd and even numbers...you never know. Since it adds on one to the sum, I think this makes a difference. Thanks!
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by IWillSurvive » Tue Nov 08, 2011 2:51 pm
Hello. I apologize - it is definitely "A". K=5 is not enough info. We do not know what the variable "n" stands for. Also, if you add (3+1))^3 = 4^3 = 64. 64/3 = 21, remainder 1. It is like this with many numbers you try...I think it turns out to be a multiple of the actual product.
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