Quick solution ?

x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT:

A) x = w
B) x > w
C) x/y is an integer
D) w/z is an integer
E) x/z is an integer

This question is testing your knowledge of the following rule:

The sum of n consecutive integers will ALWAYS be a multiple of n when n is ODD.

The sum of n consecutive integers will NEVER be a multiple of n when n is EVEN.

(You can see this with easy test cases. If you have 3 integers, say: 1, 2, and 3, then the sum is 6, which is a multiple of 3.
If you have 2 integers, say 1 and 2, then the sum is 3, which is not a multiple of 2.)

In this case, we're told that x is the sum of y consecutive integers and that y = 2z. Because z is an integer, y is EVEN. We know from the above rules that x, which is the sum of y consecutive integers, cannot be a multiple of y when y is EVEN. Therefore, x/y is NOT an integer. Answer is C.

x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT

A)x = w
B)x > w
C)x/y is an integer
D)w/z is an integer
E)x/z is an integer

w, x, y and z are all integers.
Test the SMALLEST POSSIBLE CASE.
Let z=1.

Since z=1, it must be possible that w/z and x/z are integers.
Eliminate D and E.

Since z=1, w is the sum of 1 consecutive integer, implying that w can be ANY INTEGER.
Thus, it must be possible that x=w or that x>w.
Eliminate A and B.

The correct answer is C.

Hi GMATGuruNY,

It is mentioned that w is the sum of z consecutive integers, so can we assume that z may be 1 also?

prachi18oct wrote:Hi GMATGuruNY,

It is mentioned that w is the sum of z consecutive integers, so can we assume that z may be 1 also?

Only two constraints are given for z:
1. y=2z.
2. y and z are positive integers.
Thus, it is possible that z=1 and y=2.

this question is best to solve using sample integers.

GMATGuruNY wrote:

prachi18oct wrote:Hi GMATGuruNY,

It is mentioned that w is the sum of z consecutive integers, so can we assume that z may be 1 also?

Only two constraints are given for z:
1. y=2z.
2. y and z are positive integers.
Thus, it is possible that z=1 and y=2.

Can we also solve as below:

Since y = 2z and z is integer; y = even integer
x is sum of y consecutive integers(y can be 2,4,6,8,etc)
Let y = 2, two consecutive integers can never sum to even number , hence this will never be true.
Similarly, if y = 4; n-2,n-1,n,n+1 => sum = 4n-2 ; not divisibly by 4
y = 6; n-3,n-2,n-1,n,n+1,n+2 => sum = 6n-3; not divisible by 6.

We can see a pattern as the sum can never be a multiple of y since y = even and even number of integers don't cancel out the all the +2/-2.+1/-1 part. Always one will remain, leading to non-multiple of y.

Pls let me know if this line of reasoning is ok.

Similarly, if y = 4; n-2,n-1,n,n+1 => sum = 4n-2 ; not divisibly by 4
y = 6; n-3,n-2,n-1,n,n+1,n+2 => sum = 6n-3; not divisible by 6.

Perfectly valid. This is essentially an algebraic explanation for the rule that the sum of 'n' consecutive integers won't be a multiple of 'n' when 'n' is even.

GMAT GuruNY has good method of Elimination & solving the question most quickly.

Algebraically, we could just do:

x = m + (m + 1) + ... + (m + 2z - 1)
w = k + (k + 1) + ... + (k + z - 1)

So we have
x = 2z*m + ((2z-1)(2z)/2) = 2zm + z(2z - 1)
w = zk + ((z-1)z/2)

Now we can easily test possibilities for each answer.

A:: possible if z = 1
B:: possible with almost any positive integers
C:: this would give us

x/y = (2zm + z(2z-1)) / 2z
= m + ((2z-1)/2)

But (2z - 1)/2 will never be an integer, so x/y will never be an integer. Since this works, we don't need to test D and E, and we're done.