BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

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
Join the discussion
Source: — Problem Solving |

by bobdylan » Tue May 01, 2012 4:55 am
Is the answer A ???
Join the discussion

by mathewmithun » Tue May 01, 2012 5:11 am
I think it is A. Good question :)
Join the discussion

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
Join the discussion

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.
Join the discussion

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)
Join the discussion

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.
Join the discussion

by fyllmax » Tue May 01, 2012 12:28 pm
Why is not E?
Join the discussion

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.
Join the discussion

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.
Join the discussion