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
Start free 5-day trial
Read 559 reviewsTarget Test Prep

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

A positive integer is divisible by 3 if and only if the sum

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 187
Joined: Tue Sep 13, 2016 12:46 am
by alanforde800Maximus » Thu May 10, 2018 5:08 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the six-digit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have?

A 2
B. 3
C. 4
D. 5
E 10

Source: Question Pack 1.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Thu May 10, 2018 7:57 pm
Hi alanforde800Maximux,

We're told that the six-digit integer is divisible by 3, and is of the form 1K2,K24, where K represents a digit that occurs twice. We're asked for the number of different possible values.

Using the 'rule of 3', since 1+2+2+4 = 9 (which is a multiple of 3), the value of 2K (the sum of the other two digits in the number) must ALSO be a multiple of 3. Thus, K could be 0, 3, 6 or 9. That's 4 possible values.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8211
Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members
Location: Los Angeles, CA
Thanked: 43 times

by Scott@TargetTestPrep » Mon May 14, 2018 4:04 pm
alanforde800Maximus wrote:A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. If the six-digit integer is divisible by 3, and n is of the form 1k2,k24, where k represents a digit that occurs twice, how many values could n have?

A 2
B. 3
C. 4
D. 5
E 10
We know the following:

(1 + 2 + 2 + 4 + 2k)/3 = integer

(9 + 2k)/3 = integer

Thus, k must be a multiple of 3, so it can be 0, 3, 6, or 9.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1