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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Number Properties

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 328
Joined: Thu Aug 07, 2008 5:25 pm
Location: Philadelphia
Thanked: 4 times
GMAT Score:550
by Abdulla » Sat Jul 11, 2009 3:09 pm
111
112
113
121
....
332
+ 333
______

There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. The addition problem above shows six of these integers. If all 27 of the integers were listed, what would their sum be?

A) 2,704
B) 2,990
C) 5,404
D) 5,444
E) 5,994

OA is E
Abdulla
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 472
Joined: Sun Mar 29, 2009 6:54 pm
Thanked: 56 times

by ssmiles08 » Sat Jul 11, 2009 4:07 pm
The short way of doing this is recognizing that the numbers are occurring in sets of 3, which means at least one 3 is factorable in the sum.

So the answer choice must be at least a multiple of 3. Choice E fits that puzzle.

the long way of doing this averaging out evenly set numbers.

111+112+113+121+122+123+...

112*3 + 122*3 + ...132*3

you know there are 27 numbers, you can categorize them in groups of 3 and you will have 9 sets.

so 3*(112 + 122 + 132 + 212 + 222 + 232...)

these 9 sets of numbers can be further categorized into 3 sets of 3.

3*(122*3 + 222*3 + 332*3)

3*3(122 + 222 + 332)

the final set is also an evenly spaced set, so you can further simplify it into:

3*3*(222*3) = 3*3*3(222) = 5994

so ultimately the sum has to be a multiple of 27 which 5994 is.

(E)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 1223
Joined: Thu May 01, 2008 3:29 pmFollowed by: 15 members
Location: Los Angeles, CA
Thanked: 185 times

by VP_Jim » Sat Jul 11, 2009 5:12 pm
Here's another way:

Each column is going to add up to 54. We can figure this out by seeing that since there are 27 numbers in the column, we'll have 9 of each number (27/3), so (9x1)+(9x2)+(9x3)=54.

So, now just add. The units' digit column adds to 54, so the units' digit is 4 and we carry the 5.

The tens' column also adds to 54, but we carried the 5, so it's 59. The tens' digit is 9, and we carry the 5.

The hundreds' column also adds to 54, but we carried the 5, so it's 59.

And we have:

5994

EDIT -- Not that this saves much time in this case, but we can stop at the tens' digit since there's only one answer choice that ends in 94. Could be helpful on a similar problem.
Jim S. | GMAT Instructor | Veritas Prep
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8136
Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members
Location: Los Angeles, CA
Thanked: 43 times

by Scott@TargetTestPrep » Tue Feb 27, 2018 10:48 am

There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. The addition problem above shows six of these integers. If all 27 of the integers were listed, what would their sum be?

A) 2,704
B) 2,990
C) 5,404
D) 5,444
E) 5,994
Some of the integers that have this property are 123, 111, 213, and 322. Of course, it's possible to list all 27 such integers and then add them up. However, it will be too time-consuming. Therefore, we will use a shortcut by arguing that, of these 27 integers, the digits 1, 2, and 3 each must appear in the hundreds position 9 times. Using the same logic, they each will also appear in the tens position 9 times and in the units position 9 times. Thus, the sum of these integers is:

(100 + 200 + 300) x 9 + (10 + 20 + 30) x 9 + (1 + 2 + 3) x 9

(600) x 9 + (60) x 9 + (6) x 9

(666) x 9

5,994

Answer: E

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