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# Is K^2 odd? 1) K-1 is divisable by 2 2) The sum of K

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Is K^2 odd? 1) K-1 is divisable by 2 2) The sum of K [#permalink]  11 Jun 2010, 03:43
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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
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Re: Let's discuss the answers [#permalink]  11 Jun 2010, 03:54
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.
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Re: Let's discuss the answers [#permalink]  11 Jun 2010, 04:09
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.
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Re: Let's discuss the answers [#permalink]  11 Jun 2010, 06:27
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ykaiim wrote:
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.

sjayasa wrote:
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.

When we say that number $$a$$ is divisible by number $$b$$ it means: $$a$$ and $$b$$ are integers AND $$\frac{a}{b}=integer$$.

Is K^2 odd?

(1) K-1 is divisable by 2 --> $$k-1=2n$$ --> $$k=2n+1=odd$$ --> $$k^2=odd^2=odd$$. Sufficient.

(2) The sum of K consecutive integer is divisible by K:
• If k is odd, the sum of k consecutive integers is always divisible by k. Given $$\{9,10,11\}$$, we have $$k=3=odd$$ consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If k is even, the sum of k consecutive integers is never divisible by k. Given $$\{9,10,11,12\}$$, we have $$k=4=even$$ consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

So as $$\frac{sum \ of \ k \ consecutive \ integers}{k}=integer$$, then $$k=odd$$ --> $$k^2=odd^2=odd$$. Sufficient.

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Re: Let's discuss the answers [#permalink]  11 Jun 2010, 06:34
Thanks for the clarification Bunuel!
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Re: Let's discuss the answers [#permalink]  11 Jun 2010, 06:40
Thanks Bunuel for clearing my doubt.
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Re: Let's discuss the answers   [#permalink] 11 Jun 2010, 06:40
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# Is K^2 odd? 1) K-1 is divisable by 2 2) The sum of K

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