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# Is k^2 odd?

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Is k^2 odd? [#permalink]  08 Aug 2011, 19:24
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74% (01:55) correct 26% (01:01) wrong based on 82 sessions
Is k^2 odd?

(1) k - 1 is divisible by 2.
(2) The sum of k consecutive integers is divisible by k.
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Aug 2013, 06:13, edited 1 time in total.
Renamed the topic and edited the question.
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Re: MGmat DS [#permalink]  08 Aug 2011, 20:01
How's this.

(2) The sum of k consecutive integers is divisible by k.

This can be expressed as a + (a+1) + (a+2) ... (k-1).
Group the a's together and group the numbers together to get: ka + k(k-1)/2

Given that this sum is divisible by k, the k(k-1)/2 term must be an integer.
For that to be true, k must be odd - essentially what 1) is saying.

Great question.
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Re: MGmat DS [#permalink]  08 Aug 2011, 20:05
cellydan wrote:
How's this.

(2) The sum of k consecutive integers is divisible by k.

This can be expressed as a + (a+1) + (a+2) ... (k-1).
Group the a's together and group the numbers together to get: ka + k(k-1)/2

Given that this sum is divisible by k, the k(k-1)/2 term must be an integer.
For that to be true, k must be odd - essentially what 1) is saying.

Great question.

How can u say K must be odd...even if its even k(k-1)/2 is an integer.
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Re: MGmat DS [#permalink]  08 Aug 2011, 20:25
Ah, you are right. I have omitted one additional step.

Group the a's together and group the numbers together to get: ka + k(k-1)/2
Factor out a k.
k(a+ (k-1)/2)
For this to be divisible by k, (a + (k-1)/2) must be a integer. So therefore k must be odd, otherwise the (k-1)/2) term would be a fraction...
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Re: MGmat DS [#permalink]  08 Aug 2011, 20:27
hey cellydan thanks alot for the explanation....btw when is your GMAT?
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Re: MGmat DS [#permalink]  08 Aug 2011, 23:06
ruturaj wrote:
Is k2 odd?
(1) k - 1 is divisible by 2.
(2) The sum of k consecutive integers is divisible by k.

Question Is k^2 odd?

Stmt1: k-1 is divisible by 2.

(k-1)/2 = m where m is some integer
k-1 = 2m
k = 2m+1. Now this is the equation of any odd number

Hence k is odd. Therefore k^2 is odd.
Sufficient.

Stmt2: The sum of k consecutive integers is divisible by k
sum = k(k+1)/2
Sum/k = m where m is some integer
k(k+1)/2k = m
(k+1)/2 = m
k=2m-1. This again represents odd number.

Hence k is odd. Therefore k^2 is odd.
Sufficient.

OA D
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Re: MGmat DS [#permalink]  09 Aug 2011, 02:56

I would like to point out a small possible improvement in your analysis, which is otherwise on track.

The sum of k consecutive integers is not k(k+1)/2. This is the expression for the sum of the first k natural numbers.
Example: The sum of three consecutive integers 50,51,and 52 is not 3(3+1)/2.

The expression for the sum of k consecutive integers is ak + k(k-1)/2, where a is the first number and k is the number of terms. In this question, you may have assumed that k consecutive integers can also be n consecutive natural numbers and so used the expression k(k+1)/2.

The OA is (D) as explained by cellydan
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Re: MGmat DS [#permalink]  21 Aug 2013, 05:55
Hi, What if I assume K is a fraction/decimal?

ruturaj wrote:
Is k2 odd?
(1) k - 1 is divisible by 2.
(2) The sum of k consecutive integers is divisible by k.

Question Is k^2 odd?

Stmt1: k-1 is divisible by 2.

(k-1)/2 = m where m is some integer
k-1 = 2m
k = 2m+1. Now this is the equation of any odd number

Hence k is odd. Therefore k^2 is odd.
Sufficient.

Stmt2: The sum of k consecutive integers is divisible by k
sum = k(k+1)/2
Sum/k = m where m is some integer
k(k+1)/2k = m
(k+1)/2 = m
k=2m-1. This again represents odd number.

Hence k is odd. Therefore k^2 is odd.
Sufficient.

OA D
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Re: MGmat DS [#permalink]  21 Aug 2013, 06:22
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Expert's post
ganesamurthy wrote:
Hi, What if I assume K is a fraction/decimal?

ruturaj wrote:
Is k2 odd?
(1) k - 1 is divisible by 2.
(2) The sum of k consecutive integers is divisible by k.

Question Is k^2 odd?

Stmt1: k-1 is divisible by 2.

(k-1)/2 = m where m is some integer
k-1 = 2m
k = 2m+1. Now this is the equation of any odd number

Hence k is odd. Therefore k^2 is odd.
Sufficient.

Stmt2: The sum of k consecutive integers is divisible by k
sum = k(k+1)/2
Sum/k = m where m is some integer
k(k+1)/2k = m
(k+1)/2 = m
k=2m-1. This again represents odd number.

Hence k is odd. Therefore k^2 is odd.
Sufficient.

OA D

Both statements imply that k is an integer.

Is k^2 odd?

(1) k - 1 is divisible by 2 --> $$k-1=even$$ --> $$k=even+1=odd=integer$$ --> $$k^2=odd^2=odd$$. Sufficient.

(2) The sum of k consecutive integers is divisible by k. Here k must be a positive integer because otherwise the statement does not make sense.

Properties of consecutive integers:
• If n is odd, the sum of n consecutive integers is always divisible by n. Given $$\{9,10,11\}$$, we have $$n=3=odd$$ consecutive integers. The sum is 9+10+11=30, which is divisible by 3.
• If n is even, the sum of n consecutive integers is never divisible by n. Given $$\{9,10,11,12\}$$, we have $$n=4=even$$ consecutive integers. The sum is 9+10+11+12=42, which is NOT divisible by 4.

The statement says that the sum of k consecutive integers is divisible by k, which, according to the above means that k is odd, therefore $$k^2=odd^2=odd$$. Sufficient.

For more check Number Theory chapter of our Math Book: math-number-theory-88376.html

Hope it helps.
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Re: Is k^2 odd? [#permalink]  28 Oct 2014, 04:35
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Re: Is k^2 odd?   [#permalink] 28 Oct 2014, 04:35
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