The addition problem above shows four of the 24 different in : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 21 Jan 2017, 09:10

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# The addition problem above shows four of the 24 different in

Author Message
TAGS:

### Hide Tags

Intern
Joined: 25 Oct 2010
Posts: 46
WE 1: 3 yrs
Followers: 0

Kudos [?]: 119 [3] , given: 13

The addition problem above shows four of the 24 different in [#permalink]

### Show Tags

02 Nov 2010, 23:34
3
KUDOS
13
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

74% (02:19) correct 26% (01:34) wrong based on 386 sessions

### HideShow timer Statistics

1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Nov 2012, 02:17, edited 1 time in total.
Renamed the topic and edited the question.
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 105

Kudos [?]: 958 [5] , given: 25

### Show Tags

02 Nov 2010, 23:45
5
KUDOS
3
This post was
BOOKMARKED
student26 wrote:
1,234
1,243
1,324
.....
....
+4,321

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?

A.24,000
B.26,664
C.40,440
D.60,000
E.66,660

Using the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit.

The sum is therefore = (1 + 10 + 100 + 1000) * (1*6 +2*6 +3*6 +4*6) = 1111 * 6 * 10 = 66660

_________________
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2021
Followers: 161

Kudos [?]: 1705 [5] , given: 376

### Show Tags

08 Feb 2011, 05:26
5
KUDOS
4
This post was
BOOKMARKED
1,2,3,4 can be arranged in 4! = 24 ways

The units place of all the integers will have six 1's, six 2's, six 3's and six 4's
Likewise,
The tens place of all the integers will have six 1's, six 2's, six 3's and six 4's
The hundreds place of all the integers will have six 1's, six 2's, six 3's and six 4's
The thousands place of all the integers will have six 1's, six 2's, six 3's and six 4's

Addition always start from right(UNITS) to left(THOUSANDS);

Units place addition; 6(1+2+3+4) = 60.
Unit place of the result: 0
carried over to tens place: 6

Tens place addition; 6(1+2+3+4) = 60 + 6(Carried over from Units place) = 66
Tens place of the result: 6
carried over to hunderes place: 6

Hundreds place addition; 6(1+2+3+4) = 60 + 6(Carried over from tens place) = 66
Hundreds place of the result: 6
carried over to thousands place: 6

Thousands place addition; 6(1+2+3+4) = 60 + 6(Carried over from hundreds place) = 66
Thousands place of the result: 6
carried over to ten thousands place: 6

Ten thousands place of the result: 0+6(Carried over from thousands place) = 6

Result: 66660

Ans: "E"
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36590
Followers: 7091

Kudos [?]: 93333 [3] , given: 10557

### Show Tags

08 Feb 2011, 05:48
3
KUDOS
Expert's post
17
This post was
BOOKMARKED
Merging similar topics.

Formulas for such kind of problems (just in case):

1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

Similar questions:
nice-question-and-a-good-way-to-solve-103523.html
can-someone-help-94836.html
sum-of-all-3-digit-nos-with-88864.html
permutation-88357.html
sum-of-3-digit-s-78143.html
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7125
Location: Pune, India
Followers: 2137

Kudos [?]: 13677 [2] , given: 222

### Show Tags

22 Jan 2013, 20:07
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
hellscream wrote:

Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?

The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make.

How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed?
You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit)

1111
1112
1121
... and so on till 4444

By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4.
Same for 10s, 100s and 1000s place.

Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)
= (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4)
= 1111*64*10 = 711040

or use the formula given by Bunuel above:
Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times)
=$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Manager
Joined: 01 Nov 2010
Posts: 181
Location: Zürich, Switzerland
Followers: 2

Kudos [?]: 38 [0], given: 20

### Show Tags

09 Nov 2010, 05:24
Thanks for the great explaination shrouded1!
Senior Manager
Joined: 10 Nov 2010
Posts: 267
Location: India
Concentration: Strategy, Operations
GMAT 1: 520 Q42 V19
GMAT 2: 540 Q44 V21
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 301 [0], given: 22

### Show Tags

08 Feb 2011, 04:09
the addition problem below shows four of the 24 different integers that can be formed by using each of the digits 1,2,3, and 4 exactly once in each integer.what is the sum of these 24 integers?

1,234
1,243
1,324
.......
.......
+4,321

a) 24,000
b) 26,664
c) 40,440
d) 60,000
e) 66,660
_________________

The proof of understanding is the ability to explain it.

Manager
Joined: 13 Feb 2012
Posts: 144
GMAT 1: 720 Q49 V38
GPA: 3.67
Followers: 0

Kudos [?]: 11 [0], given: 107

### Show Tags

22 Jan 2013, 09:44
Bunuel wrote:
Merging similar topics.

Formulas for such kind of problems (just in case):

1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.

Similar questions:
nice-question-and-a-good-way-to-solve-103523.html
can-someone-help-94836.html
sum-of-all-3-digit-nos-with-88864.html
permutation-88357.html
sum-of-3-digit-s-78143.html

Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
_________________

Intern
Joined: 21 May 2013
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

The addition problem above shows four of the 24 different intege [#permalink]

### Show Tags

21 May 2013, 06:35
This can be solved much easier by realizing that, since the number of four term permutations is 4!, and that summing the a sequence to its reverse gives

1234 +4321 = 5555

1243 +3421 = 5555

we may see that there are 4!/2 pairings we can make, giving us

5555(12) = 66660
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13483
Followers: 576

Kudos [?]: 163 [0], given: 0

Re: The addition problem above shows four of the 24 different in [#permalink]

### Show Tags

16 Jun 2014, 03:04
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.
_________________
Intern
Joined: 25 Jul 2014
Posts: 20
Concentration: Finance, General Management
GPA: 3.54
WE: Asset Management (Venture Capital)
Followers: 0

Kudos [?]: 23 [0], given: 52

Re: The addition problem above shows four of the 24 different in [#permalink]

### Show Tags

28 Sep 2014, 09:41
For those who could not memorize the formular, you can guess the answer in 30 secs:
Since we have 24 numbers, we will have 6 of 1 thousand something, 6 of 2 thousand something, 6 of 3 thousand something, and 6 of 4 thousand something
So,
6x1(thousand something) = 6 (thousand something)
6x2(thousand something) = 12 (thousand something)
6x3(thousand something) = 18 (thousand something)
6x4(thousand something) = 24 (thousand something)
Add them all 6+12 +18 + 24 = 60 (thousand something)
----> E
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13483
Followers: 576

Kudos [?]: 163 [0], given: 0

Re: The addition problem above shows four of the 24 different in [#permalink]

### Show Tags

11 Nov 2015, 23:06
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.
_________________
Intern
Joined: 24 Apr 2016
Posts: 6
Followers: 0

Kudos [?]: 0 [0], given: 895

Re: The addition problem above shows four of the 24 different in [#permalink]

### Show Tags

15 Aug 2016, 13:38
each term is repeating 6 times..(total 24/4=4)
now at unit...each term will repeat 6 times... (1x6+ 2x6 + 3x6 + 4x6 = 60) , so unit digit is "0" and 6 remaining.

repeating same.....total of 24 ten digit will be 60 + 6 from total of unit ,so ten digit will be 6.

only E has 60 as last two digits.
Re: The addition problem above shows four of the 24 different in   [#permalink] 15 Aug 2016, 13:38
Similar topics Replies Last post
Similar
Topics:
3 The figure above shows a side view of the insert and the four componen 3 30 Jul 2016, 05:16
1 In the addition problem above, A and B represent digits in two differe 1 11 Dec 2015, 08:22
13 The addition problem above shows four of the 24 13 20 Nov 2012, 23:53
45 There are 24 different four-digit integers than can be 18 04 Nov 2012, 09:39
3 As the figure above shows, four identical circles are inscri 3 17 Feb 2011, 14:04
Display posts from previous: Sort by

# The addition problem above shows four of the 24 different in

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.