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Number Properties from MGMAT

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by fangtray » Tue May 01, 2012 4:47 am
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

x = w
x > w
x/y is an integer
w/z is an integer
x/z is an integer
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by bobdylan » Tue May 01, 2012 4:55 am
Is the answer A ???
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by mathewmithun » Tue May 01, 2012 5:11 am
I think it is A. Good question :)
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by mathbyvemuri » Tue May 01, 2012 5:19 am
As all other options can easily be proved to be true, option 'A' is the answer
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by spartacus1412 » Tue May 01, 2012 8:04 am
let us suppose w is sum of z consecutive nos beginning form 0
and x is sum of y consecuetive nos beginning from 0

hence, x= y*(y+1)/2
hence, x/y = (y+1)/2 --eq(1)
because y = 2z given . substitute in eq(1)

we get , x/y = (2z+1)/2 which is (odd number)/2
This can never be an integer.

so C should be the answer.
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by LalaB » Tue May 01, 2012 9:05 am
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
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by mathewmithun » Tue May 01, 2012 10:32 am
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.
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by fyllmax » Tue May 01, 2012 12:28 pm
Why is not E?
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by mathewmithun » Tue May 01, 2012 7:36 pm
fyllmax wrote:Why is not E?
From eq3: x=2za+z(2z+1)

x/z=2a+(2z+1) which is definitely an integer since a is an integer and z is an integer. Properties of a and z are implied in question itself


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

So option E is definitely an integer.
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by oldsole00 » Wed May 02, 2012 6:04 am
If I understand what this question is asking, it doesn't seem to be very well written. Specifically, I'm questioning why the author would stipulate that y & z be considered positive/negative? If you're talking about consecutive integers, it's not possible to have "negative" consecutive integers, even if the set in question contains negative numbers. A better written question might specify that "x & w are the sums of y & z consecutive positive integers, respectively, where y = 2z" or something like that. If that were the question, the answer would be A.
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