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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.
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
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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.
Answer: A
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.
Answer: A
17 Aug 2006, 08:42
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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
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
