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# Is k^2 odd?

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Is k^2 odd? [#permalink]

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08 Aug 2011, 19:24
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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]

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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: MGmat DS [#permalink]

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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]

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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]

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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]

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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]

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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]

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09 Aug 2011, 02:56
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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]

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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: Is k^2 odd? [#permalink]

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Re: Is k^2 odd? [#permalink]

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25 Sep 2016, 10:27
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Re: Is k^2 odd? [#permalink]

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26 Sep 2016, 02:24
Question asks is k odd?

1. Clearly Suff, k-1 --> even so k is odd
2. Choose a case or two to test. 3,4,5 yes div by 3
3,4 not div by 2
1,2,3,4,5 - yes div by 5

So, click on D & move on

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Re: Is k^2 odd? [#permalink]

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21 Nov 2016, 07:45
Here we need to get i k is even/odd
Statement 1
(k-1)/2=integer
hence k-1=2m for some integer m
k=2m+1=> odd
hence sufficient
Statement 2
Here we can use a simple Rule =>

SUM OF N CONSECUTIVES IS DIVISIBLE BY N FOR N BEING ODD AND NEVER DIVISIBLE BY N FOR N BEING EVEN

Hence K must be odd
Alternatively,
Since consecutives inters form an AP
mean = median
It is given that mean = integer
so median => integer too
hence number of terms must be odd
hence K is odd

Thus sufficient

Hence D
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Re: Is k^2 odd? [#permalink]

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21 Nov 2016, 10:40
ruturaj wrote:
Is k^2 odd?

(1) k - 1 is divisible by 2.
(2) The sum of k consecutive integers is divisible by k.

FROM STATEMENT - I ( SUFFICIENT )

Since, k - 1 is divisible by 2 ; k must be Odd because -

Odd - 1 = Even ( Which is divisible by 2 )

FROM STATEMENT - II ( SUFFICIENT )

Test using numebrs...

Sum of 2 consecutive integers is 3 ( which is not divisible by 2 )
Sum of 3 consecutive integers is 6 ( which is divisible by 3 )
Sum of 4 consecutive integers is 10 ( which is not divisible by 4 )
Sum of 5 consecutive integers is 15 ( which is divisible by 5 )

So, we can safely conclude k = Odd...

And $$Odd^2$$ = Odd

Thus, EACH statement ALONE is sufficient to answer the question asked, answer will be (D)....

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Re: Is k^2 odd? [#permalink]

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23 Feb 2017, 03:49
PROMPT ANALYSIS
K is a natural number of the form 2n (even) or 2n+1 (odd)

SUPERSET
The answer will be either YES or NO.

TRANSLATION
In order to find the value, we need:
Exact value of k.
Any equation to solve for k.
Any characteristics of k.

STATEMENT ANALYSIS

St 1: if k is 2n, k-1 is 2n -1 which odd and hence cannot be divisible by 2. If k is 2n+1, k-1 is 2n, which is even, hence divisible by 2. Therefore k is odd. SUFFICIENT. Hence option b, c, e eliminated.

St 2: let us take the following set:a, a+1, a+2,a+3 ………. a+k -1.
Sum is equal to ka + k(k-1)/2
When sum is divided by k we get a +(k-1)/2. For the expression to be an integer, k-1 has to be even, hence k is odd. SUFFICIENT, hence option a is rejected.

Option D.

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Re: Is k^2 odd?   [#permalink] 23 Feb 2017, 03:49
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