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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]

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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
Correct answer D

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, 03:19, edited 2 times in total.
Edited the question.
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Re: Average of Consecutive Integers [#permalink]

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New post 02 Jul 2011, 14:52
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]

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New post 02 Jul 2011, 14:55
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]

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New post 03 Jul 2011, 01:02
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]

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New post 03 Jul 2011, 01:44
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]

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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]

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New post 03 Jul 2011, 02:29
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]

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New post 05 Nov 2013, 03:11
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]

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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]

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New post 31 Jan 2014, 14:24
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
Correct answer D

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]

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New post 01 Feb 2014, 04:03
+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]

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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]

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New post 26 Apr 2016, 19:24
stmt 1 says that n is even

we can try with 2 integers: 2 and 3 , the average is 2.5
we can try with 4 integers: 2, 3, 4, 5, the average is 3.5

there is a rule that the average of a consecutive even set of numbers will not result in an integer.

Sufficient.


stmt2

0 < S < n, where S = Sum of terms and n = number of terms.

S = n * average of set

manipulate this formula to Average = S/n

If S < n, then S/n cannot be equal to 1.

Sufficient.
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]

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New post 26 Apr 2016, 22:45
if average has to be = 1 , sum of elements must be equal to n . E G. 0,1,2 ; -1,0,1,2,3 .

if number of elements is odd in consecutive numbers, middle element= average.
if number of elements is even , average = sum of middle two elements/2 ---> odd number/2 so average would not be an integer.

statement 1: n is even therefore n can not be equal to 1 so it gives a definite Ans
statement 2: 0<S<n but for average to be 1, S must be equal to n so it gives a definite Ans

therefore D.
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Is the average of n consecutive integers equal to 1 ? [#permalink]

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New post 20 Dec 2016, 17:55
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Great Question.
Here is what i did in this one ->

We are asked if the mean of n consecutive is an integer or not.

N consecutive integer => Evenly spaced set =>Mean = Median =Average of the first and the last term

Mean for N consecutive integers can take only 2 forms =>
x => For N = odd
x.5=> for N=even

for some integer x


Statement 1-->
N is even

So mean will never be an integer(Because median will never be an integer)
Hence mean can never be "one".
Hence Sufficient

Statement 2-->
Sum=> S
And 0<S<N

\(Mean = \frac{Sum}{#}\)


Mean = \(\frac{S}{N}\)

As \(0<S<N\)=>

\(\frac{S}{N}\) will be a decimal between (0,1)

Hence mean will never be one.
Hence Sufficient


Hence D

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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]

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New post 15 Feb 2017, 12:43
fivedaysleft wrote:
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...


If Set = { -1,0,1,2} then Sum = 2 and n = 4, so S < n. Isn't this a valid scenario. In this case S < n S-2 is valid.
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]

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New post 15 Feb 2017, 13:36
coolkl wrote:
fivedaysleft wrote:
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...


If Set = { -1,0,1,2} then Sum = 2 and n = 4, so S < n. Isn't this a valid scenario. In this case S < n S-2 is valid.



Yes, that's a valid scenario. But is the average = 1?
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]

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New post 06 Mar 2017, 07:47
Hi Bunuel ,

Could you post the detail solution for this question .

Regards,
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]

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New post 06 Mar 2017, 10:05
abhisheknandy08 wrote:
Hi Bunuel ,

Could you post the detail solution for this question .

Regards,


Is the average of n consecutive integers equal to 1 ?

Any set of consecutive integers is an evenly spaced set, where mean = median. Thus for the average to be 1 we should have one of the following types of sets:
{1}
{0, 1, 2}
{-1, 0, 1, 2, 3}
So, basically there should be equal number of elements less than 1 and greater than 1. You can notice that in this case the number of elements turns out to be odd.

(1) n is even --> n is not odd --> 1 cannot be the average. Sufficient.

(2) if S is the sum of the n consecutive integers, then 0 < S < n --> divide by n: 0 < S/n < 1. Since S/n (the sum of the elements divided by the number of elements) is the average, then we directly got that the average is not 1, it's < 1. Sufficient.

Answer: D.

Hope it's clear.
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Re: Is the average of n consecutive integers equal to 1 ?   [#permalink] 06 Mar 2017, 10:05
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