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You are given a series of n consecutive positive integers

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You are given a series of n consecutive positive integers [#permalink] New post 11 Jul 2003, 18:40
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You are given a series of n consecutive positive integers where n > 1. Is the average value of the 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.

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
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Re: composed problem #6 [#permalink] New post 11 Jul 2003, 18:50
AkamaiBrah wrote:
You are given a series of n consecutive positive integers where n > 1. Is the average value of the 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.


(1) tells nothing on its own
(2) (n-1)/2 cannot be an integer unless (n-1) is even, i.e., unless n is odd. So let us only experiment with an odd n:

4,5,6 => 4+ (3-1)/2 = 4+1 = 5 => not divisible by 3
2,3,4 => 2+ (3-1)/2 = 2+1 = 3 => divisible by 3

So (2) on its own is not helpful either.

However, it was established that (1) is actually a pre-existing condition that must be met in order to hold (2) fully true. So combining the statements will be of no help, and we must choose (E).
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Re: composed problem #6 [#permalink] New post 11 Jul 2003, 20:42
JP wrote:
AkamaiBrah wrote:
You are given a series of n consecutive positive integers where n > 1. Is the average value of the 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.


(1) tells nothing on its own
(2) (n-1)/2 cannot be an integer unless (n-1) is even, i.e., unless n is odd. So let us only experiment with an odd n:

4,5,6 => 4+ (3-1)/2 = 4+1 = 5 => not divisible by 3
2,3,4 => 2+ (3-1)/2 = 2+1 = 3 => divisible by 3

So (2) on its own is not helpful either.

However, it was established that (1) is actually a pre-existing condition that must be met in order to hold (2) fully true. So combining the statements will be of no help, and we must choose (E).


Nice try, but E is NOT correct.
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 [#permalink] New post 11 Jul 2003, 22:27
Looking at it again, I can't say how I'd do it differently. Sorry.
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 [#permalink] New post 11 Jul 2003, 23:15
I will go with B

THe average of the series is [n/2(2a +(n-1)*1]/n i.e a + (n-1)/2

since in statement 2 it is given that a+(n-1)/2 is divisible by 3, it implies that that average is divisible by 3
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 [#permalink] New post 12 Jul 2003, 01:31
The "trick" to solve DS problems is to "restate" the conditions in a way that is relevent to question we need to answer. (Sometimes, you have to restate the question stem into a direct question also).

(1) restated says that the average is the middle number of the series (the average of an odd number of consecutive tems is always the middle term), but with no other information, we have no idea if it is divisible by 3.

(2) as brstorewala correctly points out, is just a fancy way to say that the AVERAGE of the series is an integer divisible by 3. Since that directly answers the question, it is sufficient. (proof: let K be the first number. K + (n-1) is the last number. the average of the two = average of the series = K + (n-1)/2 )

Hence, B is the correct answer.
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Re: composed problem #6 [#permalink] New post 10 Sep 2008, 14:14
Answer should be B

Let 1st term be a and the nth term be n. In a consecutive series, nth term is a +(n-1) in value. According to the question, a + [a +(n-1) - 1]/2 is divisible by 3.
This implies, (3a + n - 2)/2 /3 = Integer, say I.
If we break up the numerator, we get (a + n)/2 /3 + (2a -2)/2 /3 = I
i.e., average/3 + (a-1)/3 + I
Therefore, the average value of the series is, infact divisible by 3.
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Re: composed problem #6 [#permalink] New post 10 Sep 2008, 23:34
tricky one.

I tried to use numbers to prove and messed up resulting in E as the answer.
Re: composed problem #6   [#permalink] 10 Sep 2008, 23:34
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