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Consecutive integers

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Source: — Data Sufficiency |
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by cans » Sun Jun 05, 2011 8:10 am
odd+even=odd
thus z1,z2,...zn
s=z1+z2+....+zn
s is odd??
a)s/n is odd -> s=n*odd number. If n is even, s is even, if n is odd, s is odd. Insufficient
b)n is odd. insufficient. (2,3,4 sum->odd ; 3,4,5 sum->odd)
a&b) n is odd and thus sum is odd. Sufficient
IMO C
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by finance » Sun Jun 05, 2011 8:14 am
Thank you Cans! I made the same mistake as you, but that's not the OA.
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by [email protected] » Sun Jun 05, 2011 8:26 am
If z1,z2......zn is a series of consecutive positive integers, is the sum of all the integers in this sequence odd?

(1) (z1+z2+....zn)/n is an odd integer
(2) n is odd


Statement 1 says that it is an arithmetic mean of all given values and it is odd but alone is not sufficient.

Statement 2 gives the value of n but the value of 'n' alone is not sufficient.

Both statements together will give an answer i.e odd number X odd number = odd number
eg: 7 X 9 = 63.
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by cans » Sun Jun 05, 2011 8:36 am
finance wrote:Thank you Cans! I made the same mistake as you, but that's not the OA.
found my mistake
zn=z1+(n-1)
s=(n/2)*(z1+zn)
s/n = z1 + (n-1)/2
z1 is integer. also s/n is odd integer.
Thus (n-1)/2 must be integer. or n-1 is even
or n is odd.
Thus A alone is sufficient (s=odd)
IMO A
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by Anurag@Gurome » Sun Jun 05, 2011 8:40 am
finance wrote:If z1,z2......zn is a series of consecutive positive integers, is the sum of all the integers in this sequence odd?

(1) (z1+z2+....zn)/n is an odd integer
(2) n is odd
Say, x is the first integer of the series.
Then the sum of the all integers = x + (x + 1) + (x + 2) + ... + (x + n - 1) = nx + n(n - 1)/2 = n(x + (n - 1)/2)

Statement 1: (z1 + z2 + ... + zn)/n is an odd integer
Hence, (x + (n - 1)/2) is an odd integer.
Hence, (n - 1)/2 is an integer, i.e. n is an odd integer.
Therefore, sum of all the integers in the sequence = n*(some odd integer) = odd*odd = odd

Sufficient.

Statement 2: n is odd.
Hence, sum of the integers = odd*(x + (n - 1)/2)
But we don't know whether x is odd or even. Also we don't know whether (n - 1)/2 is even or odd.

Not sufficient.

The correct answer is A.
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by aftableo2006 » Mon Jun 06, 2011 9:03 pm
thanks for the explanation this was a tricky one
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by gmat1978 » Wed Jun 08, 2011 10:34 pm
Cans,

Can you please explain your solution below in more detail?

Thanks
cans wrote:
finance wrote:Thank you Cans! I made the same mistake as you, but that's not the OA.
found my mistake
zn=z1+(n-1)
s=(n/2)*(z1+zn)
s/n = z1 + (n-1)/2
z1 is integer. also s/n is odd integer.
Thus (n-1)/2 must be integer. or n-1 is even
or n is odd.
Thus A alone is sufficient (s=odd)
IMO A
Great question.
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by cans » Wed Jun 08, 2011 11:12 pm
gmat1978 wrote:Cans,

Can you please explain your solution below in more detail?

Thanks
cans wrote:
finance wrote:Thank you Cans! I made the same mistake as you, but that's not the OA.
found my mistake
zn=z1+(n-1)
s=(n/2)*(z1+zn)
s/n = z1 + (n-1)/2
z1 is integer. also s/n is odd integer.
Thus (n-1)/2 must be integer. or n-1 is even
or n is odd.
Thus A alone is sufficient (s=odd)
IMO A
Great question.
as z1,z2,...,zn is series of consecutive integers,
z2=z1+1
z3=z2+1 = z1+2
thus zn = z1 + (n-1)
Sum = z1 + z1 + z3 + ....... + zn
also using sum of arithmetic progression series formula
sum = (n/2)(z1+zn) (sum = no. of terms * (first term + last term) / 2)
sum = (n/2)(z1+z1+(n-1))
sum=(n/2) (2z1 + (n-1))
sum/n = (2z1 + (n-1))/2
also sum/n is odd.
Thus (2z1 + (n-1))/2 is odd
multiplying that odd number with 2 will give even number
thus 2z1 + (n-1) is even
also 2z1 is even
thus (n-1) + even = even -> (n-1) has to be even
thus n is even+1 or n is odd
also given sum/n is odd
Thus sum=n*odd
as n is odd, sum=odd*odd = odd (ex: 5*3=15; 3*9=27)
Sufficient
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