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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.

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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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?
_________________

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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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

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

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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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01 Oct 2015, 03:50

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

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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

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

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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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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?
_________________

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.

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