Last visit was: 16 Jul 2024, 02:27 It is currently 16 Jul 2024, 02:27
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# There are 27 different three-digit integers that can be formed using

SORT BY:
Tags:
Show Tags
Hide Tags
Intern
Joined: 02 Jan 2011
Posts: 2
Own Kudos [?]: 157 [157]
Given Kudos: 0
Manager
Joined: 02 Apr 2010
Posts: 78
Own Kudos [?]: 258 [47]
Given Kudos: 18
GMAT Club Legend
Joined: 08 Jul 2010
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Posts: 6020
Own Kudos [?]: 13789 [39]
Given Kudos: 125
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
General Discussion
Intern
Joined: 24 Jun 2010
Posts: 9
Own Kudos [?]: 14 [8]
Given Kudos: 0
Re: There are 27 different three-digit integers that can be formed using [#permalink]
5
Kudos
3
Bookmarks
here you go: 27 digits are formed by 3 * 3 *3 that is..1st place can take 1,2,3...again 2nd place can take all three digits...3rd place can also take all three digits.

now lets consider the last digit as 1 out of the three that we can take:

3*3*(1)

so with 1 as the last digit we can have 9 combinations.so we have 9 1's in the last place

similary,lets consider the last digit as 2 out of the three that we can take:

3*3*(2)

so with 2 as the last digit we can have 9 combinations.so we have 9 2's in the last place

similary,lets consider the last digit as 3 out of the three that we can take:

3*3*(3)

so with 3 as the last digit we can have 9 combinations.so we have 9 3's in the last place

therefore sum of 9 1's,9 2's,9 3's will be 9+18+27=54...similarly 2nd and 1st place will give u 54...therefore 5994..hope u got it!!
Manager
Joined: 13 Jun 2012
Posts: 166
Own Kudos [?]: 508 [6]
Given Kudos: 467
Location: United States
WE:Supply Chain Management (Computer Hardware)
Re: There are 27 different three-digit integers that can be formed using [#permalink]
2
Kudos
4
Bookmarks
The simplest way is this formula n^n-1 *sum of the numbers (111)= 5994. where n= total number, which is 3 here. so 3^3-1=9. 9+6(111)=5994
Manager
Joined: 10 Mar 2013
Posts: 136
Own Kudos [?]: 497 [0]
Given Kudos: 2412
GMAT 1: 620 Q44 V31
GMAT 2: 610 Q47 V28
GMAT 3: 700 Q49 V36
GMAT 4: 690 Q48 V35
GMAT 5: 750 Q49 V42
GMAT 6: 730 Q50 V39
GPA: 3
Re: There are 27 different three-digit integers that can be formed using [#permalink]
swaraj wrote:
here you go: 27 digits are formed by 3 * 3 *3 that is..1st place can take 1,2,3...again 2nd place can take all three digits...3rd place can also take all three digits.

now lets consider the last digit as 1 out of the three that we can take:

3*3*(1)

so with 1 as the last digit we can have 9 combinations.so we have 9 1's in the last place

similary,lets consider the last digit as 2 out of the three that we can take:

3*3*(2)

so with 2 as the last digit we can have 9 combinations.so we have 9 2's in the last place

similary,lets consider the last digit as 3 out of the three that we can take:

3*3*(3)

so with 3 as the last digit we can have 9 combinations.so we have 9 3's in the last place

therefore sum of 9 1's,9 2's,9 3's will be 9+18+27=54...similarly 2nd and 1st place will give u 54...therefore 5994..hope u got it!!

This is a great explanation! Thanks!
Intern
Joined: 18 Dec 2015
Posts: 1
Own Kudos [?]: 11 [6]
Given Kudos: 1
GPA: 3.15
WE:Analyst (Non-Profit and Government)
Re: There are 27 different three-digit integers that can be formed using [#permalink]
5
Kudos
1
Bookmarks
How I did it:

123 is a possible solution, which is divisible by 3. E is the only option that adds up to a number divisible by 3.
GMAT Club Legend
Joined: 19 Dec 2014
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Posts: 21835
Own Kudos [?]: 11783 [3]
Given Kudos: 450
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: There are 27 different three-digit integers that can be formed using [#permalink]
2
Kudos
1
Bookmarks
Hi All,

This question has a great series of built-in 'logic shortcuts' that don't require a lot of advanced math (although a bit of arithmetic will still be required to get to the correct answer).

We're told that there are 27 3-digit numbers that can be formed with the digits 1, 2 and 3 (the smallest being 111 and the largest being 333). Without too much effort, you should be able to figure out that...

9 of the numbers begin with a '1'
9 of the numbers begin with a '2'
9 of the numbers begin with a '3'

So when we add up JUST the 100s, we have.... 9(100) + 9(200) + 9(300) = 5400

Next, consider ALL of the 10s that we have... 27 of them (and some are 20s and some are 30s). That would add HUNDREDS (well over 270+) to the total sum. Based on how the answer choices are written, there's only one answer that makes sense...

GMAT assassins aren't born, they're made,
Rich
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 19133
Own Kudos [?]: 22642 [13]
Given Kudos: 286
Location: United States (CA)
Re: There are 27 different three-digit integers that can be formed using [#permalink]
6
Kudos
7
Bookmarks
rdhookar wrote:
There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. 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

Manager
Joined: 30 Jul 2014
Status:MBA Completed
Affiliations: IIM
Posts: 91
Own Kudos [?]: 97 [1]
Given Kudos: 107
Re: There are 27 different three-digit integers that can be formed using [#permalink]
1
Kudos
Sum of the numbers should be divisible by 9 as each number is appearing 9 times. Divisibility rule by 9 conforms(Sum of all the digits should be divisible by 9) only to the answer option E - hence answer should be E.
Manager
Joined: 01 Dec 2018
Posts: 146
Own Kudos [?]: 146 [6]
Given Kudos: 333
Concentration: Entrepreneurship, Finance
Schools: HBS '21 ISB'22
GPA: 4
WE:Other (Retail Banking)
Re: There are 27 different three-digit integers that can be formed using [#permalink]
5
Kudos
1
Bookmarks
rdhookar wrote:
There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. 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

SUPER EASY METHOD that always works

STEP 1 always find the total no. of combinations possible = here we have been given 27 (use fundamental method to check/find 3x3x3 =27)
STEP 2 - find the average of lowest and highest number ( 123 + 321 = 444 . 444/2 = average 222)
STEP 3- multiply the average with the possible numbers . = 27 * 222 = 5994 .
Manager
Joined: 26 Apr 2019
Posts: 247
Own Kudos [?]: 168 [0]
Given Kudos: 135
Location: India
GMAT 1: 690 Q49 V34
GMAT 2: 700 Q49 V36
GMAT 3: 720 Q50 V37
GMAT 4: 740 Q50 V40
GPA: 3.99
Re: There are 27 different three-digit integers that can be formed using [#permalink]
at ones place there will be nine 1 , nine 2 and nine 3 , adding all 27 will give you 54 . silmilarly you will get 54 at tens place and 54 at hundred place . all all we get e as answer
VP
Joined: 07 Dec 2014
Posts: 1067
Own Kudos [?]: 1607 [0]
Given Kudos: 27
Re: There are 27 different three-digit integers that can be formed using [#permalink]
rdhookar wrote:
There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. 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

111+333=444
444/2=222
222*27=5994
E
Intern
Joined: 20 Jul 2020
Posts: 6
Own Kudos [?]: 1 [0]
Given Kudos: 27
Re: There are 27 different three-digit integers that can be formed using [#permalink]
I think there is a simple formula that can be applied here.

My only confusion is regarding how we can figure out whether or not the question implies repetition of numbers.

The formula without the repetition gives the answer of course, but can we determine that for sure from the question?
GMAT Club Legend
Joined: 19 Dec 2014
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Posts: 21835
Own Kudos [?]: 11783 [0]
Given Kudos: 450
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: There are 27 different three-digit integers that can be formed using [#permalink]
salphonso wrote:
I think there is a simple formula that can be applied here.

My only confusion is regarding how we can figure out whether or not the question implies repetition of numbers.

The formula without the repetition gives the answer of course, but can we determine that for sure from the question?

Hi salphonso,

The prompt specifically tells us that we're dealing with "27 DIFFERENT three-digit integers...", so we know that the numbers do NOT repeat. GMAT questions are always carefully worded, so you should pay attention to the specific 'math vocabulary' that is used in each prompt (as that can help you to properly 'restrict' your thinking or get you thinking about possibilities that are not immediately obvious to you).

GMAT assassins aren't born, they're made,
Rich
Intern
Joined: 20 Jul 2020
Posts: 6
Own Kudos [?]: 1 [0]
Given Kudos: 27
Re: There are 27 different three-digit integers that can be formed using [#permalink]
Hi Rich,

Can't believe I missed that! Thank you!
Intern
Joined: 18 Jul 2017
Posts: 11
Own Kudos [?]: 4 [0]
Given Kudos: 212
There are 27 different three-digit integers that can be formed using [#permalink]
Can we use this method?

1 number would be 111 and last would be 333.
Their average will be 222.
Using the average method- 222*27 is 5994.

Bunuel

Posted from my mobile device
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6804
Own Kudos [?]: 30813 [2]
Given Kudos: 799
Re: There are 27 different three-digit integers that can be formed using [#permalink]
2
Kudos
Top Contributor
rdhookar wrote:
There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. 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

This is a great example of how the GMAT often rewards students for thinking outside the box.

Here, we can apply a divisibility rule that says integer N is divisible by 3 if and only if the sum of the digits of N is divisible by 3.
For example, we know that 11112 is divisible by 3, because 1+1+1+1+5=9, and 9 is divisible by 3.

Notice that 1+2+3 = 6, and 6 is divisible by 3
This means any 3-digit integer consisting of a 1, a 2 and a 3 must be divisible by 3
If each of the 27 integers in the sum is divisible by 3, then the sum must also be divisible by 3.
In other words, the correct answer must be divisible by 3, which means the sum of its digits must be divisible by 3.

A. 2+7+0+4 = 13, which is not divisible by 3. Eliminate.
B. 2+9+9+0 = 20, which is not divisible by 3. Eliminate.
C. 5+4+0+4 = 13, which is not divisible by 3. Eliminate.
D. 5+4+4+4 = 17, which is not divisible by 3. Eliminate.
E. 5+9+9+4 = 27, which IS divisible by 3.

By the process of elimination, the correct answer is E.
Non-Human User
Joined: 09 Sep 2013
Posts: 33988
Own Kudos [?]: 851 [0]
Given Kudos: 0
Re: There are 27 different three-digit integers that can be formed using [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: There are 27 different three-digit integers that can be formed using [#permalink]
Moderator:
Math Expert
94355 posts