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Re: Series of n consecutive positive integers [#permalink]
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]
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]
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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Given a series of n consecutive positive integers, where n > [#permalink]
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.


(1)
consider 2+3+4+5+6+7+8 =35
also, n= odd.

consider 2+3+4 = 9


(2)

according to second statement, 2+((7-1)/2) = 8


but for n=9, 2+ ((9-1)/2) = 6

So here you go, answer is E
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Re: Given a series of n consecutive positive integers, where n > [#permalink]
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Re: Given a series of n consecutive positive integers, where n > [#permalink]
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