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29 Aug 2004, 04:57
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30 Aug 2004, 00:36
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k& = k^2+3
k& = 4d1+3 = k^2+3
k^2 = 4d1
So k^2 must be a multiple of 4, from 1, it can be 20,24,28,32,36
But we're also told k^2 is an integer, so k^2 must be able to be sqaure-root into a whole number. Of the numbers above only 36 is possible, so k is 6.
To check, 6& = 36+3 = 39
39/4 = 6(4) + R3
So (1) is sufficient.
From (2), we're told k is a factor of 24, so k can be 2, 3, 4, 6, 8, 12, 24
But
2& = 7, and 7/4 = 4(1) + R3
3& = 12 and 12/4 = 3(4) + R0 -- out
4& = 19 and 19/4 = 4(4) + R2 -- out
6& = 39 and 39/4 = 4(6) + R3 -- so we have more than one value of k that can give us a remainder of 3 when k& is divided by 4
Therefore (2) is not sufficient and the answer should be (A)
k& = 4d1+3 = k^2+3
k^2 = 4d1
So k^2 must be a multiple of 4, from 1, it can be 20,24,28,32,36
But we're also told k^2 is an integer, so k^2 must be able to be sqaure-root into a whole number. Of the numbers above only 36 is possible, so k is 6.
To check, 6& = 36+3 = 39
39/4 = 6(4) + R3
So (1) is sufficient.
From (2), we're told k is a factor of 24, so k can be 2, 3, 4, 6, 8, 12, 24
But
2& = 7, and 7/4 = 4(1) + R3
3& = 12 and 12/4 = 3(4) + R0 -- out
4& = 19 and 19/4 = 4(4) + R2 -- out
6& = 39 and 39/4 = 4(6) + R3 -- so we have more than one value of k that can give us a remainder of 3 when k& is divided by 4
Therefore (2) is not sufficient and the answer should be (A)
