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11 Jun 2010, 04:43
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Is K^2 odd?
1) K-1 is divisable by 2
2) The sum of K consecutive interger is divisable by K
1) K-1 is divisable by 2
2) The sum of K consecutive interger is divisable by K
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11 Jun 2010, 04:54
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IMO B or D.
1. If K = 2\(\sqrt{2}\)+1, then
k-1 is divisible by 2, but \(k^2\) is not odd.---------------I am not sure if this is correct as a square root is divisible by an integer. If Yes, then S1 is sufficient.
2. S2 is Sufficient.
1. If K = 2\(\sqrt{2}\)+1, then
k-1 is divisible by 2, but \(k^2\) is not odd.---------------I am not sure if this is correct as a square root is divisible by an integer. If Yes, then S1 is sufficient.
2. S2 is Sufficient.
11 Jun 2010, 05:09
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I am not very sure too. But IMO D. I guess if you say K is divisible by 2, then the multiple should not be irrational.
A alone is SUFFICIENT. IMO
B alone is SUFFICIENT. For any odd number K, the sum of the first K consecutive numbers is always divisible by K.
Proof for B:
Any odd numnber K can be represented as 2n+1(where n is an integer).
The sum of any 2n+1 consecutive integers, with a as the first term and 2n+1 as the last term, is;
a + (a+d) + (a+2d) + (a+3d) + ... + (a+(n-1)d) = n(first term + last term)/2
a + (a+1) + (a+2) + ... + (a+2n) = (2n+1)(2a+2n)/2 = (2n+1)(a+n) [common difference d = 1 in this case]
It is obvious that this is divisible by 2n+1, our original odd number.
A alone is SUFFICIENT. IMO
B alone is SUFFICIENT. For any odd number K, the sum of the first K consecutive numbers is always divisible by K.
Proof for B:
Any odd numnber K can be represented as 2n+1(where n is an integer).
The sum of any 2n+1 consecutive integers, with a as the first term and 2n+1 as the last term, is;
a + (a+d) + (a+2d) + (a+3d) + ... + (a+(n-1)d) = n(first term + last term)/2
a + (a+1) + (a+2) + ... + (a+2n) = (2n+1)(2a+2n)/2 = (2n+1)(a+n) [common difference d = 1 in this case]
It is obvious that this is divisible by 2n+1, our original odd number.
