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Number Properties from MGMAT

Expert replies
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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Source: — Problem Solving |

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.
Its do or die this time!
Practise, practise and practise.
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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
Happy are those who dream dreams and are ready to pay the price to make them come true.(c)

In order to succeed, your desire for success should be greater than your fear of failure.(c)
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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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