If y is the sum of x consecutive positive integers, and x3, is y odd?

Question

If y is the sum of x consecutive positive integers, and x>3, is y odd?

(1) x is odd.
(2) The largest number among the x consecutive integers is odd.

sumitkrocks · Answer

y =  \frac{(Ist term + Last term )}{2}  * x

say terms are 
Case 1: 1 to 7 
x=7 terms , sum= y = 28 ; y is even
 
Case 2: 1 to 5
 x=5 terms , sum= y = 15 ; y is odd

Case 1 and 2 both meet statement 1 and 2 conditions with different out come

(E)  is the answer IMO

kantapong · Answer

Trick: Push the question as further as you can!
y is the sum of x consecutive integers => Suppose y starts at n: y = n + (n+1) + ... + (n+(x-1)) = nx + (x-1)x/2
Is y odd? => is nx + (x-1)x/2 odd? => is nx odd? => are both n and x odd?

1) x is odd (not sufficient to answer whether both n and x are odd)
2) largest number is odd => n + x - 1 is odd => n + x is even
Case 1: n is even and x is even => Ans = no
Case 2: n is odd and x is odd => Ans = yes (insufficient)
3) Combine both: x is odd (1) => n is odd (2) => both n and x are odd (sufficient)
Ans = C

Archit3110 · Answer

target determine whether sum of consecutive + integers 'y' is odd
x is count of integers and x>3

#1
 x is odd
case 1: ( 1,2,3,4,5) ; y is odd ; Yes to target
case 2: ( 2,3,4,5,6) ; y is even ; No to target
insufficient
#2
The largest number among the x consecutive integers is odd.
case 1: ( 1,2,3,4,5) ; y is odd ; Yes to target
case 2: ( 1,2,3,4,5,6,7) ; y is even ; no to target
from 1 &2
we can have 
case 1: ( 1,2,3,4,5) ; y is odd ; Yes to target
case 2: ( 1,2,3,4,5,6,7) ; y is even ; no to target
insufficient
option E

HarshaBujji · Answer

IMO E

As per the given condition, all the numbers are consecutive. o,e,o,e.... (Odd and Even)
So sum of these x numbers depends on the : Whether x is odd/even:

Lets see ;
If x is even then the possibilties are {e,o,e,o} => Sum is e or {e,o,e,o,e,o } => Sum os o

If x is odd {e,o,e,o,e} =>Sum is e, {o,e,o,e,o} Sum is o {e,o,e,o,e,o,e} => Sum is o, {o,e,o,e,o,e,o}=> Sum is e

Hence we cannot conclude this even after using both statements.

Bunuel · Answer

If y is the sum of x consecutive positive integers, and x>3, is y odd?

The sum of x consecutive integers equals the mean multiplied by the number of terms, thus x*{mean}=y. Recall that in evenly spaced set mean=median, thus we have that x*{mean}=x*{median}=y.

The question asks whether y is odd.

(1) x is odd --> x*{median}=odd*{median}=y. If a set has odd number of terms the median is the middle term, when arranged in ascending/descending order, thus the median of the set must be an integer. For y to be odd the median must be odd, but the median of odd number of consecutive integers can be odd as well as even, for example, consider {1, 2,  3 , 4, 5} and {1, 2, 3,  4 , 5, 6, 7}. Not sufficient.

(2) The largest number among the x consecutive integers is odd. If the set is {0,  1, 2,  3}, then y is even but if the set is {1, 2,  3 , 4, 5}, then y is odd. Not sufficient.

(1)+(2) The sets we considered to prove insufficiency of the first statement are still valid because both of them have the largest term odd (to satisfy the second statement). Thus even taken together the statements are not sufficient.

Answer: E.

Kinshook · Answer

Asked: If y is the sum of x consecutive positive integers, and x>3, is y odd?

(1) x is odd.
If x=5; y = 1+2+3+4+5 = 15; y is odd
But if x=7; y= 1+2+3+4+5+6+7 = 28; y is even
NOT SUFFICIENT

(2) The largest number among the x consecutive integers is odd.
If largest number=5; y = 1+2+3+4+5 = 15; y is odd
But if largest number=7; y= 1+2+3+4+5+6+7 = 28; y is even
NOT SUFFICIENT

(1) x is odd.
(2) The largest number among the x consecutive integers is odd.
If x=5; y = 1+2+3+4+5 = 15; y is odd
But if x=7; y= 1+2+3+4+5+6+7 = 28; y is even
NOT SUFFICIENT

IMO E

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If y is the sum of x consecutive positive integers, and x>3, is y odd?

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If y is the sum of x consecutive positive integers, and x>3, is y odd?

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