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# Is the average of n consecutive integers equal to 1 ?

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Is the average of n consecutive integers equal to 1 ? [#permalink]  02 Jul 2011, 13:30
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Is the average of n consecutive integers equal to 1 ?

(1) n is even
(2) if S is the sum of the n consecutive integers, then 0 < S < n

[Reveal] Spoiler: My approach

Hello Fluke and Everyone in this forum - can you please explain the conceptual way of to solve this and improve my understanding ?

My approach to solve this:

Statement 1: Sufficient, because the average of the even number of consecutive integers can never be an integer.

Statement 2: I understand what the statement means, but I am struggling to understand the concept.

My understanding is

Average = Sum/Number of terms
Therefore, Sum = Average * Number of Terms
Therefore S=Average/n. But what can we conclude from this? The book says that this is sufficient as well.

So the answer is D. But I don't understand the Statement 2.
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 Nov 2013, 02:19, edited 2 times in total.
Edited the question.
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Re: Average of Consecutive Integers [#permalink]  02 Jul 2011, 13:52
statement 1 is fairly clear as you said

statement 2 says : S is positive and S>n.....(eq)
now as you said sum = avg*n = S

subtitute value of S in equation we get => avg*n>n
now since n cant be negative, we can cancel n out on both sides without a change of sign
which gives me =>avg>1 so it succesfully answers NO! to is avg=1?

hope its clear...
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Re: Average of Consecutive Integers [#permalink]  02 Jul 2011, 13:55
enigma123 wrote:
Is the average of $$n$$ consecutive integers equal to $$1$$?

1) $$n$$ is even
2) If $$S$$ is the sum of the $$n$$ consecutive integers then $$0<S>n$$.

2) To get an average=1; the Sum of elements must be equal to number of elements.

$$If S=n; \frac{S}{n}=1$$;

If S is NOT equal to n, the average can not be 1.

Sum=100; n=100; S/n=100/100=1
Sum=1; n=1; S/n=1/1=1
Sum=10; n=9; S/n=10/9=1.##(NOT EQUAL TO 1)

Sufficient.
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Re: Average of Consecutive Integers [#permalink]  03 Jul 2011, 00:02
My apologies. The second statement should have read

if S is the sum of the n consecutive integers, then 0<S<n.
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Re: Average of Consecutive Integers [#permalink]  03 Jul 2011, 00:44
So if 0<s<n, then also the answer is average cannot be 1, because s has at least to be 1 and n hs to be greater than 1, so in either case the average cannot be 1. Is my thinking correct fluke and everyone please?
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Re: Average of Consecutive Integers [#permalink]  03 Jul 2011, 01:27
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enigma123 wrote:
So if 0<s<n, then also the answer is average cannot be 1, because s has at least to be 1 and n hs to be greater than 1, so in either case the average cannot be 1. Is my thinking correct fluke and everyone please?

s<n or s>n means $$s \ne n$$. If s NOT EQUAL n, the average can't be 1.
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Re: Average of Consecutive Integers [#permalink]  03 Jul 2011, 01:29
Many thanks Fluke. Appreciate all your help.
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Re: Is the average of n consecutive integers equal to 1 ? 1) n [#permalink]  05 Nov 2013, 02:11
Dear enigma123,
Please edit stm2 as you posted it will be "if S is the sum of the n consecutive integers, then 0<S<n".
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Re: Is the average of n consecutive integers equal to 1 ? 1) n [#permalink]  05 Nov 2013, 02:22
Expert's post
monirjewel wrote:
Dear enigma123,
Please edit stm2 as you posted it will be "if S is the sum of the n consecutive integers, then 0<S<n".

Edited the original post.
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]  31 Jan 2014, 13:24
enigma123 wrote:
Is the average of n consecutive integers equal to 1 ?

(1) n is even
(2) if S is the sum of the n consecutive integers, then 0 < S < n

[Reveal] Spoiler: My approach

Hello Fluke and Everyone in this forum - can you please explain the conceptual way of to solve this and improve my understanding ?

My approach to solve this:

Statement 1: Sufficient, because the average of the even number of consecutive integers can never be an integer.

Statement 2: I understand what the statement means, but I am struggling to understand the concept.

My understanding is

Average = Sum/Number of terms
Therefore, Sum = Average * Number of Terms
Therefore S=Average/n. But what can we conclude from this? The book says that this is sufficient as well.

So the answer is D. But I don't understand the Statement 2.

Is the average of n consecutive integers = 1?

Statement 1

n is even

If n is even then the average of consecutive integers will not be an integer at all so it won't be 1 for sure

Just to illustrate

n = 2, the avarege of 1 and 2 is 1.5
n=4, the average of 1,2,3 and 4 is 2.5 etc...

Sufficient

Statement 2

We can manipulate this to 0<s/n<1

So we are told that the average is in fact less than 1

Therefore sufficient as well

D
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]  01 Feb 2014, 03:03
+1 D

Stmt1: n is even.

let n=4, and 4 consecutive integers can be a-1,a,a+1,a+2. sum=a-1+a+a+1+a+2 = 4a+2 their average is (4a+2)/4 a+1/2 since a is integer and adding a fraction to a. mean can never be 1. SUFF

Stmt2: 0<S<n.

for average to be 1 S must be equal to n. Hence Stmt2 is also SUFF.
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Re: Is the average of n consecutive integers equal to 1 ?   [#permalink] 01 Feb 2014, 03:03
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