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# Is the sum of the series of six consecutive integers even?

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Is the sum of the series of six consecutive integers even? [#permalink]  16 Nov 2005, 19:27
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Is the sum of the series of six consecutive integers even?

1) the first integer in the series is even
2) the six digit number formed by the series is a multiple of nine.
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alrighty then...lets see..

(1) is simpilest...pick 0 as the starting...you get odd sum..try -2. try 2...you get odd number...suff

(2) well if the (9m-5)+(9m-4)...(9m)...well if m is positive number which it is..give us a definite number...

D it is..
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I say D.

A is sufficient because the sum of 6 consecs that start with a even number will always be odd.

b. is sufficient because I picked the numbers 2,3,4,5,6,7 which add up to 27- a multiple of 9. once again we know it's odd
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I think it is A,

2 cannot give you an answer, consider 1,2,3,4,5,6 or 3,4,5,6,7,8 both are not divisible by 9 whereas 2,3,4,5,6,7 is divisible and hence we cannot say conclusively.
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D is the OA.
But I am unsure how stmt 2 plays out if you have -ve consecutive integers. There isn't really a proper OE but here's my reasoning:

Well, the sum S=6n+15=3(2n+5), where n is the first number

stmt 1) irrespective of the what n is, S will be always be odd. Suff.

stmt 2) says that the 6 digit number formed by each integer is divisible by 9 --> meaning the sum of the numbers is divisible by 9 --> 3(2n+5) = 9x
therefore, (2n+5)=3x . How do we conclude from this whether the sum is even or odd?
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Re: DS-consecutive numbers [#permalink]  17 Nov 2005, 02:52
tank wrote:
Is the sum of the series of six consecutive integers even?

how can a series of 6 consecutive numbers not be odd ?
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Re: DS-consecutive numbers [#permalink]  17 Nov 2005, 09:46
christoph wrote:
tank wrote:
Is the sum of the series of six consecutive integers even?

how can a series of 6 consecutive numbers not be odd ?

yes..if we take 6 consecutive integers as:

n-2 n-1 n n+1 n+2 n+3

sum: 6n + 3
for all intger n, 6n will be even and 6n+3 be odd.
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tank wrote:
D is the OA.
But I am unsure how stmt 2 plays out if you have -ve consecutive integers. There isn't really a proper OE but here's my reasoning:

Well, the sum S=6n+15=3(2n+5), where n is the first number

stmt 1) irrespective of the what n is, S will be always be odd. Suff.

stmt 2) says that the 6 digit number formed by each integer is divisible by 9 --> meaning the sum of the numbers is divisible by 9 --> 3(2n+5) = 9x
therefore, (2n+5)=3x . How do we conclude from this whether the sum is even or odd?

Tank....

If you let 6n+15 equal to a multiple of 9, you will find out that the first term is an even number in the series. And from (1), we know that the sum is always odd. So (2) is sufficient.
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