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Is k^2 odd? [#permalink]
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08 Aug 2011, 20:24
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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.
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Last edited by Bunuel on 21 Aug 2013, 07:13, 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, 07:22
ganesamurthy wrote: Hi, What if I assume K is a fraction/decimal? jamifahad wrote: 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: k1 is divisible by 2. (k1)/2 = m where m is some integer k1 = 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=2m1. 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 > \(k1=even\) > \(k=even+1=odd=integer\) > \(k^2=odd^2=odd\). 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. Answer: D. For more check Number Theory chapter of our Math Book: mathnumbertheory88376.htmlHope it helps.
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Re: MGmat DS [#permalink]
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08 Aug 2011, 21:01
How's this.
(2) The sum of k consecutive integers is divisible by k.
This can be expressed as a + (a+1) + (a+2) ... (k1). Group the a's together and group the numbers together to get: ka + k(k1)/2
Given that this sum is divisible by k, the k(k1)/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, 21:05
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) ... (k1). Group the a's together and group the numbers together to get: ka + k(k1)/2
Given that this sum is divisible by k, the k(k1)/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(k1)/2 is an integer.



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Re: MGmat DS [#permalink]
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08 Aug 2011, 21:25
Ah, you are right. I have omitted one additional step.
Group the a's together and group the numbers together to get: ka + k(k1)/2 Factor out a k. k(a+ (k1)/2) For this to be divisible by k, (a + (k1)/2) must be a integer. So therefore k must be odd, otherwise the (k1)/2) term would be a fraction...



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Re: MGmat DS [#permalink]
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08 Aug 2011, 21:27
hey cellydan thanks alot for the explanation....btw when is your GMAT?



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Re: MGmat DS [#permalink]
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09 Aug 2011, 00:06
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: k1 is divisible by 2. (k1)/2 = m where m is some integer k1 = 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=2m1. 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, 03:56
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jamifahad, 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(k1)/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, 06:55
Hi, What if I assume K is a fraction/decimal? jamifahad wrote: 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: k1 is divisible by 2. (k1)/2 = m where m is some integer k1 = 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=2m1. 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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25 Sep 2016, 11:27
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Re: Is k^2 odd? [#permalink]
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26 Sep 2016, 03:24
Question asks is k odd?
1. Clearly Suff, k1 > 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, 08:45
Here we need to get i k is even/odd Statement 1 (k1)/2=integer hence k1=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, 11:40
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, 04:49
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, k1 is 2n 1 which odd and hence cannot be divisible by 2. If k is 2n+1, k1 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(k1)/2 When sum is divided by k we get a +(k1)/2. For the expression to be an integer, k1 has to be even, hence k is odd. SUFFICIENT, hence option a is rejected.
Option D.











