DS, Series, Number Properties

[bretania]

Guys, do you have an easy and consistent explanation or observation as to why the answer here is A?

If Z1, Z2, Z3, … Zn is a series of consecutive positive integers, is the sum of all the integers in this series odd?

(1) (Z1 + Z2 + Z3 + … +Zn)/n is an odd integer
(2) n is odd

[clock60]

Originally Posted by bretania

Guys, do you have an easy and consistent explanation or observation as to why the answer here is A?

If Z1, Z2, Z3, … Zn is a series of consecutive positive integers, is the sum of all the integers in this series odd?

(1) (Z1 + Z2 + Z3 + … +Zn)/n is an odd integer
(2) n is odd

i am not sure in general approach but lets try anymore
first if you are given the series of consecutive positive numbers be sure that it is arithmitic series (1 2 3 4 ........N)
the sum of can be find
Sn=((a1+an)/2)*n ( if seen closely it is a mean*number of terms)
for checking 4 5 6 7 8 mean of series 6 or (4+8)/2=6
in order your Z1+........Zn to be odd, mean and number of terms must be both odd.
(1) st directly says that mean is odd ( z1+.......zn)/n is other way to find mean (sum of terms/number of terms)
but what about n?
again in order mean of consecutive series be integer (in our case it is odd integer) n must be odd
for example (12 13 14 15 16 17) the number of term is even, mean is (14+15)/2=14,5
but if the number of terms will be odd the mean is middle number so integer
thus n is odd so the sum is also odd
may be too wordy but now is the best i can imagine
other approaches will be more helpful i am sure

[Dr_GMAT]

I like your approach

[bretania]

Clock60, thanks to this by the way.

[yuvakris]

Let me put my approach here.

Let the sum of Z1, Z2, Z3, … Zn be (z1+z2+z3...+zn)= z(1+2+3+4..n)
--> z((n*(n+1))/2). The way to find the sum of consecutive numbers is (n*(n+1))/2, where n is the last number.

Now as per the first statement (Z1 + Z2 + Z3 + … +Zn)/n is odd. i.e z*n*(n+1)/2n= odd, so we an easily find it out that the sum is odd.

[shailender.bisht]

sum of all integer in series = n/2 [ 2z1 +n-1]
= n [ z1 + (n-1)/2]

now according to (1) [z1 +(n-1)/2 ] is an odd integer

=> one of z1 and (n-1)/2 is odd ( odd +even =odd, odd+odd=even ,even+even=even)

but (n-1) is always even else (n-1)/2 qill not be an integer => n =1 +even =odd

so n is Odd and [z1 +(n-1)/2 ] is Odd

so odd *odd =odd