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 Your Progress

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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

03 Nov 2010, 00:34

3

This post received KUDOS

9

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

74% (02:20) correct
26% (01:35) wrong based on 288 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?

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

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.

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

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: \(n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)\).

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: \(n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)\).

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? _________________

Kudos!!!... If you think I help you in some ways....

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.

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) _________________

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

Show Tags

21 May 2013, 07:35

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

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

Show Tags

16 Jun 2014, 04:04

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. _________________

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

Show Tags

28 Sep 2014, 10:41

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

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

Show Tags

12 Nov 2015, 00:06

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. _________________

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...