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# Given a series of n consecutive positive integers, where n >

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Given a series of n consecutive positive integers, where n > [#permalink]

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12 Jan 2012, 23:35
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Given a series of n consecutive positive integers, where n > 1, is the average value of this series an integer divisible by 3?

(1) n is odd
(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3

For me the answer is D. can someone please let me know if you think its not correct? Unfortunately, I don't have an OA. This is how I solved it. Also, please let me know if there is any shortcut or any concept that you guys can see straight after reading the questions.

Statement 1 --> n is ODD. When there are ODD terms in the consecutive series the average will be an integer that will never be divisible by 3. Hence NO and therefore this statement is sufficient.

Statement 2 --> Let's say n=7 and 1st term = 3. So 3+(7-1/2) = 6. 6+3 = 9 which is divisible by 3. Average value will be 6 which is divisible by 3.

Lets say n = 13. Average value will be 9 which is divisible by 3. Therefore this statement is also sufficient to answer the question.
[Reveal] Spoiler: OA

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Re: Series of n consecutive positive integers [#permalink]

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13 Jan 2012, 04:19
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The sum of n terms of a consecutive positive integer series = (n/2)[2a + n -1] where a is the first term

Average of the first n terms = (1/2) [2a+ n -1] = a + [(n-1)/2]

Using statement (1), if n is odd then the answer may or may not be divisible by 3. For example, take a=1 and n=5 to get average of first n terms as 3, which is divisible by 3. However, if a=1 and n=3 then the average of the first n terms is 2, which is not divisible by 3. Insufficient.

Using statement (2), a + [(n-1)/2] is divisible by 3. This means that a + [(n-1)/2] is divisible by 3. Therefore the average is divisible by 3. Sufficient.

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Re: Series of n consecutive positive integers [#permalink]

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13 Jan 2012, 04:45
GyanOne wrote:
The sum of n terms of a consecutive positive integer series = (n/2)[2a + n -1] where a is the first term

Average of the first n terms = (1/2) [2a+ n -1] = a + [(n-1)/2]

Using statement (1), if n is odd then the answer may or may not be divisible by 3. For example, take a=1 and n=5 to get average of first n terms as 3, which is divisible by 3. However, if a=1 and n=3 then the average of the first n terms is 2, which is not divisible by 3. Insufficient.

Using statement (2), a + [(n-1)/2] is divisible by 3. This means that a + [(n-1)/2] is divisible by 3. Therefore the average is divisible by 3. Sufficient.

Here how did you take n=1, in the question he said n>1 right?...can you please explain??
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Re: Series of n consecutive positive integers [#permalink]

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13 Jan 2012, 23:40
For Statement II:

lets assume n=11 and the consecutive positive integer series be 1,2,3,4,5,6,7,8,9,10,11

then 1+(11-1)/2 = 6 which is divisible by 3 and the average of (1+2+3+4+5+6+7+8+9+10+11) =33 which is divisible by 3
however, if we assume n=3 and take the consecutive positive integer series to be 2,3,4
the 2+(3-1)/2= 3 which is divisible by 3 but the average of (2+3+4) is not divisible by 3

I'm not sure if what im doing is right ;@
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Re: Series of n consecutive positive integers [#permalink]

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16 Jan 2012, 18:05
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enigma123 wrote:
Given a series of n consecutive positive integers, where n > 1, is the average value of this series an integer divisible by 3?
(1) n is odd
(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3

For me the answer is D. can someone please let me know if you think its not correct? Unfortunately, I don't have an OA. This is how I solved it. Also, please let me know if there is any shortcut or any concept that you guys can see straight after reading the questions.

Statement 1 --> n is ODD. When there are ODD terms in the consecutive series the average will be an integer that will never be divisible by 3. Hence NO and therefore this statement is sufficient.

Statement 2 --> Let's say n=7 and 1st term = 3. So 3+(7-1/2) = 6. 6+3 = 9 which is divisible by 3. Average value will be 6 which is divisible by 3.

Lets say n = 13. Average value will be 9 which is divisible by 3. Therefore this statement is also sufficient to answer the question.

Responding to a pm.

Couple of things:
In any evenly spaced set (AP) the arithmetic mean (average) is equal to the median and can be calculated by the formula: (first term+last term)/2

Now, set of consecutive integers is an evenly spaced set (AP with common difference of 1) and in order mean=median to be an integer it has to have an odd number of terms. If there are an even number of terms mean=median will be integer/2.

For example:
{1, 2, 3} --> mean=median (middle term)=(3+1)/2=2;
{1, 2, 3, 4} --> mean=median=(1+4)/2=5/2=2.5.

Next, in AP if the first term is $$a$$ and the common difference of successive members is $$d$$, then the $$n_{th}$$ term of the sequence is given by: $$a_ n=a+d(n-1)$$. In case of consecutive integers (so when common difference=d=1) the formula becomes: $$a_ n=a+n-1$$.

Check Number Theory for more on AP: math-number-theory-88376.html (specifically "Consecutive Integers" and "Evenly Spaced Set" chapters of it)

BACK TO THE ORIGINAL QUESTION.
Given a series of n consecutive positive integers, where n > 1, is the average value of this series an integer divisible by 3?

Basically we are asked whether: $$average=\frac{first \ term+last \ term}{2}$$ is divisible by 3 or whether $$average=\frac{a+(a+n-1)}{2}=a+\frac{n-1}{2}$$ is divisible by 3 (where $$a$$ is the first term and $$(a+n-1)$$ is the $$n_{th}$$, so last term).

(1) n is odd --> as n=odd then the average is definitely an integer, though it may or may not be divisible by 3: {2, 3, 4} - YES, {1, 2, 3} - NO. Not sufficient.

(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3 --> we are directly given that $$a+\frac{n-1}{2}$$ is divisible by 3. Sufficient.

Important note: you should have spotted that there was something wrong with your solution as on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.

So we can not have answer NO from statement (1) and answer YES from statement (2) (as you got in your solution), because in this case statements would contradict each other.

Hope it helps.
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Re: Series of n consecutive positive integers [#permalink]

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19 Jan 2012, 16:54
You are a star Bunuel. Thanks for such a detailed explanation.
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Re: Given a series of n consecutive positive integers, where n > [#permalink]

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24 Oct 2014, 08:58
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Re: Given a series of n consecutive positive integers, where n > [#permalink]

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25 Oct 2014, 01:34
enigma123 wrote:
Given a series of n consecutive positive integers, where n > 1, is the average value of this series an integer divisible by 3?

(1) n is odd
(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3

For me the answer is D. can someone please let me know if you think its not correct? Unfortunately, I don't have an OA. This is how I solved it. Also, please let me know if there is any shortcut or any concept that you guys can see straight after reading the questions.

Statement 1 --> n is ODD. When there are ODD terms in the consecutive series the average will be an integer that will never be divisible by 3. Hence NO and therefore this statement is sufficient.

Statement 2 --> Let's say n=7 and 1st term = 3. So 3+(7-1/2) = 6. 6+3 = 9 which is divisible by 3. Average value will be 6 which is divisible by 3.

Lets say n = 13. Average value will be 9 which is divisible by 3. Therefore this statement is also sufficient to answer the question.

B.

(1) n is odd
1,2,3 ; Ans = No.
2,3,4 ; Ans = Yes.
So insufficient.

(2) The sum of the first number of the series and (n – 1) / 2 is an integer divisible by 3
Let the series be a,a+1,a+2,...,a+(n-1)
acc to FS2 [a + (n-1)/2] mod 3 = 0
the avg of series is: na + ((n-1)(n)/2) = [n (a + (n-1)/2)]
since [a + (n-1)/2] mod 3 = 0
=> n*[a + (n-1)/2] is also div by 3.
hence, the avg is div by 3.
sufficient.
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Re: Given a series of n consecutive positive integers, where n > [#permalink]

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22 May 2016, 04:35
Hello from the GMAT Club BumpBot!

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Re: Given a series of n consecutive positive integers, where n >   [#permalink] 22 May 2016, 04:35
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