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# What is the remainder when k^2 is divided by 8? 1). When k

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What is the remainder when k^2 is divided by 8? 1). When k [#permalink]

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17 Aug 2006, 08:20
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What is the remainder when k^2 is divided by 8?
1). When k is divided by 2, the remainder is 1
2). When k is divided by 3, the remainder is 2

Tried doing the usual methodology..

k=2x+1--> K^2= 4x^2+4x+1 couldnt come up with any conclusions. Had more success plugging in numbers.
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17 Aug 2006, 08:36
Required r = ? in
k^2= 8m+r
where m is some integer.

S1: k = 2m+1
=> k^2 = 4m^2 + 4m+1
=> k^2 = 4m(m+1) +1

If m = odd, m+1 = even => the product 4m(m+1) is divisible by 8

If m = even, m+1 = odd, => the product 4m(m+1) is divisible by 8

Therefore, Sufficient

S2: k=3m+2
k^2 = 3m(m+4) +4

If m= odd, m+4 = odd, not divisible by 8

If m = even, m+4 = even, then maybe divisible by 8 if m >2

Not sufficient.

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17 Aug 2006, 08:42
IMO E
used the same methodology
stmt 1:
(2k+1)^2 -->4k^2+4k+1
k=-1 remainder = 1
k=0 remainder =1
k=-1/2 remainder =0 - insuff

stmt 2:
3k+2 same thing
k=1 remainder 1
k=0 remainder 4 insuff

Am i doing this right ??? any help
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17 Aug 2006, 08:43
Hi Haas

The official answer according to the question bank is D. But I love how you attacked the problem
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17 Aug 2006, 08:46
I don't undertstand how S2 can satisfy by itself.

Do you have OE?

apollo168 wrote:
Hi Haas

The official answer according to the question bank is D. But I love how you attacked the problem
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17 Aug 2006, 08:47
Sorry my mistake. I recheck the OA is A. My bad sorry
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17 Aug 2006, 08:50
No problem.. Thanks for confirming.

apollo168 wrote:
Sorry my mistake. I recheck the OA is A. My bad sorry
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17 Aug 2006, 09:08
A

St1: k = 2x+1
then k^2 = 4x^2 + 1 + 4x
= 4x(x+1) + 1
Since x(x+1) is an even number so 4x(x+1) will be divisible by 8. So remainder will be 1.: SUFF

St2: k = 3y+2
k^2 = 9y^2 + 4 + 12y
= 3y(3y+4)+4
Only interpretation is when k^2 is divided by 3 then remainder will be 4. No conclusion can be drawn about division by 8.: INSUFF
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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17 Aug 2006, 09:20
You guys are really great. Learn so much from this site
17 Aug 2006, 09:20
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