LalaB wrote:since y = 2z, then y is even
let y=2
x is the sum of 2 consecutive integers .let them be 1 and 2, then x=3
x/y=3/2 is not an integer
C is the answ
I still think it is A and I think LalaB and Spartacus1412 got C as answer because you considered x and w as sum of first y and z consecutive numbers, but that is not the case.
Given: x is the sum of y consecutive integers. w is the sum of z consecutive integers.
This means x=a+(a+1)+(a+2).....(a+y-1) terms, a be the starting integer.
Similarly w= b+(b+1)+(b+2).....(b+z-1) terms.
therefore x=ay+y(y-1)/2...(eq1) and w=zb+z(z-1)/2...(eq2) since y=2z, substituting in eq1 we have
x=2za+z(2z+1)...(eq3).
from options: x could be greater than w depending of a,b y and z
x/y can be an integer from eq1
w/z can be integer from eq2
x/z is an integer from eq3
Now if x=w then comparing the terms we get z(2z+1)=z(z+1)/2 and we get either z=0 or z=-1/3
Since z = integer z cannot be -1/3
If z=0, then w and z are sum of z=0 consecutive integers which makes the question itself absurd. So hence x=w cannot happen. Hence A.
