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## Am I missing something?

This topic has 1 expert reply and 4 member replies
satish.twincities Newbie | Next Rank: 10 Posts
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22 Nov 2006
Posted:
2 messages

#### Am I missing something?

Sun Jan 28, 2007 6:13 pm
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

GMAT prep says (1) is sufficent to answer the question.

I wonder how????

### GMAT/MBA Expert

Stacey Koprince GMAT Instructor
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Wed Jan 31, 2007 5:25 pm
To distribute:
(k+1)^3 = (k+1)(k+1)(k+1)
let's take the first two sets of parentheses first:
(k+1)(k+1) = k^2 + 2k + 1
now, multiple in our third (k+1):
(k+1)(k^2 + 2k + 1) =
k^3 + 2k^2 + k + k^2 + 2k + 1) =
k^3 + 3k^2 + 3k + 1 = n

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

= k^2 + 3k + 3 + 1/k. In this algebraic representation of division, the numerator of the fraction is the remainder. So the remainder is always 1.

You can also try some real numbers and see if there's a pattern. We know they're both positive integers and k>1.
Try k = 2. Then n = (2+1)^3 = 27. 27/2 = 13 R1
Try k = 3. Then n = (3+1)^3 = 64. 64/3 = 21 R1
So the first thing I notice is that, when k is even, n is odd, and vice versa. The second thing I notice is that I've gotten R1 for both. Interesting.
k = 4. Then n = (4+1)^3 = 125. 125/4 = 31 R1. R1 again.
k = 5. Then n = (5+1)^3 = 216. 216/5 = 43 R1. R1 yet again!
At this point I'm feeling pretty comfortable that there's a pattern. I'm going to get R1 no matter what I try for k.

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rnschmidt Newbie | Next Rank: 10 Posts
Joined
07 Nov 2006
Posted:
9 messages
Mon Jan 29, 2007 12:33 pm
will you please explain the math for your solution jay?

jayhawk2001 Community Manager
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Sun Jan 28, 2007 7:35 pm
N = (k+1)^3 = k^3 + 3k^2 + 3k + 1

So, N/K always yields the remainder 1

Hence (A)

800guy Master | Next Rank: 500 Posts
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Sun Jan 28, 2007 7:50 pm
jayhawk2001 wrote:
N = (k+1)^3 = k^3 + 3k^2 + 3k + 1

So, N/K always yields the remainder 1

Hence (A)
yep, i got the same approach as jayhawk.

aim-wsc Legendary Member
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Sun Jan 28, 2007 9:08 pm
good analytical approach FULL MARKS 8)

sigh
i hv forgotten all .. i used other approach like any number that is just greater than one than the other number is not divisible by it and the remainder is always one.
then we have condition A

good

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