Data Suff - If the average of this set of consecutive intege

[swatikhanwalkar]

hi,
Consider this Data Sufficiency problem:
Is k*k odd?
(1) k - 1 is divisible by 2.
(2) The sum of k consecutive integers is divisible by k.
Statement (1) tells us that k - 1 is even. Therefore, k is odd, so k2 will be odd.
SUFFICIENT.
Statement (2) tells us that the sum of k consecutive integers is divisible by k. Therefore, this
sum divided by k is an integer. Moreover, the sum of k consecutive integers divided by k is
the average (arithmetic mean) of that set of k integers. As a result, Statement (2) is telling us
that the average of the k consecutive integers is itself an integer:

If the average of this set of consecutive integers is an integer, then k must be odd.

Can someone prove how the above marked in Bold mathematically?

I cannot find an example to prove this wrong, but need some explanation as to why the statement will always stand true. .

thanks

[LazyNK]

Hey Swati,
The simplest way to visualize is as follows :

First thing you need to convince yourself is that if I arrange the terms of a set in ascending or descending order, then there will be two terms in the center of the set if the number of terms "n" is even ( those terms would be (n/2)th and (n/2+1)th terms). Also, in case the number of terms "n" is odd, then there will be only one term in the center ( that term will be ((n+1)/2)th term). eg. 6 terms -> 3d and 4th terms in center and 7 terms -> 4th term in the center.

For the case of a set with even number of consecutive integers (say "n") . If we arragnge the terms of the series in ascending order, there will be two center terms, say "c" and "c+1".
Thus the series (arranged in ascending order) is as follows :
c-(n/2-1), c-(n/2-2), ...., c-2, c-1, c, c+1, c+2, ... , c+(n/2-1), c+(n/2)
(Note that all the terms in the above series are integers as n is even and hence n/2 is an integer).
Sum of the above terms= c-(n/2-1) + c-(n/2-2) + ... + c-2 + c-1 + c + c+1 + c+2 + c+3 + ... + c+(n/2-1) + c+(n/2)
= n*c + ( -(n/2-1) - (n/2-2) + ... - 2 - 1 + 0 + 1 + 2 + ... + (n/2-1) + (n/2) )
In the parenthesized terms above, pair the terms equally spaced from 0 on left and right.
-> Sum = n*c + ( [-(n/2-1) + (n/2-1)] + [-(n/2-2) + (n/2+2)] + ... + [-2 + 2] + [-1 + 1] + (n/2))
Note the term (n/2) in bold which doesn't have a corresponding equivalent term to the left of 0.
Thus, Sum = n*c + ( 0 + 0 + ... + 0 + 0 + n/2)
= n*c + n/2
So, average of the terms of the above set = (n*c+n/2)/n = c + 1/2 . Since c is an integer, c+1/2 is not an integer.
-> Average of a set of even number of consecutive integers is not an integer.

For the case of a set with odd number of consecutive integers (say "n") . If we arragnge the terms of the series in ascending order, there will be only one center term, say "c".
Thus the series (arranged in ascending order) is as follows :
c-((n+1)/2-1), c-((n+1)/2-2), ...., c-2, c-1, c, c+1, c+2, ..., c+((n+1)/2-2), c+((n+1)/2-1)
(Note that all the terms in the above series are integers as n is odd and hence (n+1)/2 is an integer).
Sum of the above terms= c-((n+1)/2-1) + c-((n+1)/2-2) + ... + c-2 + c-1 + c + c+1 + c+2 + ... + c+((n+1)/2-1) + c+((n+1)/2)
= n*c + ( -((n+1)/2-1) - ((n+1)/2-2) + ... - 2 - 1 + 0 + 1 + 2 + ... + ((n+1)/2-1) + ((n+1)/2) )
In the parenthesized terms above, pair the terms equally spaced from 0 on left and right.
-> Sum = n*c + ( [-((n+1)/2-1) + ((n+1)/2-1)] + [-((n+1)/2-2) + ((n+1)/2+2)] + ... + [-2 + 2] + [-1 + 1] )
Note that the pairing is complete above, and there is no single isolated term remaining, as was the case when "n" was even.
Thus, Sum = n*c + ( 0 + 0 + ... + 0 + 0 )
= n*c
So, average of the terms of the above set = (n*c)/n = c . Since c is an integer, the average is also an integer.
-> Average of a set of odd number of consecutive integers is an integer.
-NK

[swatikhanwalkar]

thank you for your help. . this makes sense now. .

[jnelson0612]

Great! :-)

[krishnan.anju1987]

I think the simplest way to find whether 2) is sufficient is to plug in numbers

(1+2+3)/3=2

(1+2+3+4)/4=10/4

1+2+3+4+5/5=3

hence only if k is odd, it will be divisible by k

[jnelson0612]

I think the simplest way to find whether 2) is sufficient is to plug in numbers

(1+2+3)/3=2

(1+2+3+4)/4=10/4

1+2+3+4+5/5=3

hence only if k is odd, it will be divisible by k

True, and plugging in numbers reveals the theorem: the sum of an odd number of integers is always divisible by the number of integers.