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.

When positive integer \(n\) is divided by 3, the remainder

Expert replies
by swerve » Tue May 21, 2019 12:12 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

When positive integer \(n\) is divided by 3, the remainder is 1. When \(n\) is divided by 7, the remainder is 5. What is the smallest positive integer \(p\), such that \((n+p)\) is a multiple of 21?

A. 2
B. 2
C. 5
D. 19
E. 20

The OA is B

Source: Veritas Prep
Join the discussion
Source: — Problem Solving |

by Jay@ManhattanReview » Tue May 21, 2019 10:00 pm
swerve wrote:When positive integer \(n\) is divided by 3, the remainder is 1. When \(n\) is divided by 7, the remainder is 5. What is the smallest positive integer \(p\), such that \((n+p)\) is a multiple of 21?

A. 1
B. 2
C. 5
D. 19
E. 20

The OA is B

Source: Veritas Prep
Given that the positive integer \(n\) is divided by 3, and the remainder is 1, we have n, one among 4, 7, 13, 16, 19, 22 ...

Also, given that the positive integer \(n\) is divided by 7, and the remainder is 5, we have n, one among 12, 19, 26, ...

The smallest common value for n is 19.

Since n = 19 divided by 21 leaves a remainder of 19, for it to be divisible by 21, we need to add 2 to it. Thus, the smallest positive integer \(p\), such that \((n+p)\) is a multiple of 21 is 2.

The correct answer: B

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: GRE Manhattan | ACT Tutoring San Antonio | GRE Prep Courses Boston | Houston IELTS Tutoring | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.
Join the discussion

by Scott@TargetTestPrep » Thu May 23, 2019 4:12 pm
swerve wrote:When positive integer \(n\) is divided by 3, the remainder is 1. When \(n\) is divided by 7, the remainder is 5. What is the smallest positive integer \(p\), such that \((n+p)\) is a multiple of 21?

A. 2
B. 2
C. 5
D. 19
E. 20

The OA is B

Source: Veritas Prep
Since when n is divided by 3, the remainder is 1, n could be:

1, 4, 7, 10, 13, 16, 19, 22, ...

Since when n is divided by 7, the remainder is 5, n could be:

5, 12, 19, 26, ...

We can see that n = 19 satisfies both division/remainder criteria. And if p = 2, we have n + p = 19 + 2 = 21, which is a multiple of 21.

Alternate Solution:

Since the remainder from the division of n by 3 is 1, we can express n as n = 3k + 1 for some integer k.

Similarly, n = 7s + 5 for some integer s.

We see that n + 2 = 3k + 3 = 7s + 7 is divisible by both 3 and 7; therefore it must be divisible by 21 as well. So, the smallest such integer is 2.

Answer: A and B (Note the multiple 2 among answer choices)

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
Join the discussion