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.

n is a positive integer. Is n(n+1)(n+2)/4 an even integer?

Expert replies
by Max@Math Revolution » Thu Dec 20, 2018 11:27 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

[Math Revolution GMAT math practice question]

n is a positive integer. Is n(n+1)(n+2)/4 an even integer?

1) n is an even integer
2) 1238 ≤ n ≤ 1240
Join the discussion
Source: — Data Sufficiency |

by deloitte247 » Sun Dec 23, 2018 8:48 am
$$check\ if\ \ \frac{n\left(n+1\right)\left(n+2\right)}{4}$$
n (n+1) (n+2) is the product of three consecutive integers.
The product of three consecutive integers is always divisible by 2 and 3 because the product of k consecutive integers is always divisible by k!

Statement 1
n is an even integer.
if n is an even integer, then n (n+1) (n+2) will be product of three consecutive integers that will be divisible by the multiple of 2
if n = 2

$$\frac{2\left(2+1\right)\left(2+2\right)}{4}$$
$$\frac{2\cdot3\cdot4}{4}=\frac{24}{4}=6$$
Hence, n (n+1) (n+2) will always be even if n is an even integer ;
Statement 1 is INSUFFICIENT.

Statement 2
$$1238\le n\le1240$$
$$hence,\ n\le1239$$
$$1239\ is\ a\ multiple\ of\ \ 3,\ hence\ n\left(n+1\right)\left(n+2\right)\ is\ divisible\ by\ 4$$
$$\frac{1239\left(1239+1\right)\left(1239+2\right)}{4}$$
$$\frac{1239\left(1240\right)\left(1241\right)}{4}$$
$$\frac{1906622760}{4}$$
$$476655690\ which\ is\ an\ even\ integer\ $$
statement 2 is INSUFFICIENT.

$$answer\ is\ Option\ D$$
Join the discussion

by Max@Math Revolution » Sun Dec 23, 2018 6:06 pm
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Asking for n(n+1)(n+2)/4 to be an even integer is equivalent to asking for n(n+1)(n+2) to be a multiple of 8. If n is an even integer, n and n+2 are consecutive even integers and a product of two consecutive even integers is a multiple of 8. Thus, condition 1) is sufficient.

Condition 2)
If n = 1238, n(n+1)(n+2)=1238*1239*1240 is a multiple of 8 since 1240 is a multiple of 8.
If n = 1239, n(n+1)(n+2)=1239*1240*1241 is a multiple of 8 since 1240 is a multiple of 8.
If n = 1240, n(n+1)(n+2)=1240*1241*1242 is a multiple of 8 since 1240 is a multiple of 8.
Thus, condition 2) is sufficient.

Therefore, the answer is D.
Answer: D

Note: This question is a CMT4(B) question. Condition 1) is easy to understand and condition 2) is hard. When one condition is easy to understand, and the other is hard, D is most likely to be the answer.
Join the discussion