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 350,000 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 [#permalink]
20 Nov 2012, 23:53

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

81% (02:18) correct
19% (01:39) wrong based on 62 sessions

1,257 1,275 1,527 ........ ........ +7,521

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,5 and 7 exactly once in each integer. What is the sum of the 24 integers ?

Re: The addition problem above shows four of the 24 [#permalink]
21 Nov 2012, 02:10

2

This post received KUDOS

Numbers with 1 in the thousands postion = 6 --> 6*1000 = 6000 Numbers with 2 in the thousands postion = 6 --> 6*2000 = 12000 Numbers with 5 in the thousands postion = 6 --> 6*5000 = 30000 Numbers with 7 in the thousands postion = 6 --> 6*7000 = 42000 Numbers with 1 in the hundreds postion = 6 --> 6*100 = 600 Numbers with 2 in the hundreds postion = 6 --> 6*200 = 1200 Numbers with 5 in the hundreds postion = 6 --> 6*500 = 3000 Numbers with 7 in the hundreds postion = 6 --> 6*700 = 4200 Numbers with 1 in the tens postion = 6 --> 6*10 = 60 Numbers with 2 in the tens postion = 6 --> 6*20 = 120 Numbers with 5 in the tens postion = 6 --> 6*50 = 300 Numbers with 7 in the tens postion = 6 --> 6*70 = 420 Numbers with 1 in the ones postion = 6 --> 6*1 = 6 Numbers with 2 in the ones postion = 6 --> 6*2 = 12 Numbers with 5 in the ones postion = 6 --> 6*5 = 30 Numbers with 7 in the ones postion = 6 --> 6*7 = 42

Re: The addition problem above shows four of the 24 [#permalink]
21 Nov 2012, 03:31

This is the way i solved it :

Total no. of different combos = 24, total digits 4 therefore each digit must repeat 24/4 = 6 times in each row .. So the right most row would add up to 1x6 + 2x6 + 5x6 + 7x6 = 6+12+30+42 = 90 .. Each row would add up to 90, so 90 in the first means we have 9 that carries over and we get 0 , the second time its 90+9 and 9 stays and one 9 goes to the row to the left, so the last two digits of the SUM should be 90 (E) .. We could go on and solve the exact number but since only one answer choice has the last digits as 90 we needn't go any further.. _________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: The addition problem above shows four of the 24 [#permalink]
21 Nov 2012, 06:38

vomhorizon wrote:

1,257 1,275 1,527 ........ ........ +7,521

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,5 and 7 exactly once in each integer. What is the sum of the 24 integers ?

STEP 1: Total possible 4 digit integers 4*3*2*1 = 24 Each digit (1,2,5,7) at each place(ones,tens,hundreds,thousands) repeats 4 times. so 24/4 = 6 (Each digit will repeat 6 times in each place)

STEP 2: Sum of individual distinct digits (1,2,5,7) = 15

STEP 3: sum of individual distinct digits * No of times each digit got repeated at each place = 15*6 =90

STEP 4 : As the question is based on 4 digit integers 1111 * 90 = 99990

If it was 3 digit number calc 111 * 90 = 9990

Hope this helps you...

-- Shan _________________

GMAT - Practice, Patience, Persistence Kudos if u like

Re: The addition problem above shows four of the 24 [#permalink]
21 Nov 2012, 06:44

Quote:

Another possible method....

STEP 1: Total possible 4 digit integers 4*3*2*1 = 24 Each digit (1,2,5,7) at each place(ones,tens,hundreds,thousands) repeats 4 times. so 24/4 = 6 (Each digit will repeat 6 times in each place)

STEP 2: Sum of individual distinct digits (1,2,5,7) = 15

STEP 3: sum of individual distinct digits * No of times each digit got repeated at each place = 15*6 =90

STEP 4 : As the question is based on 4 digit integers 1111 * 90 = 99990

If it was 3 digit number calc 111 * 90 = 9990

Hope this helps you...

-- Shan

Same as the way i did it... _________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: The addition problem above shows four of the 24 [#permalink]
21 Nov 2012, 08:38

vomhorizon wrote:

Quote:

Another possible method....

STEP 1: Total possible 4 digit integers 4*3*2*1 = 24 Each digit (1,2,5,7) at each place(ones,tens,hundreds,thousands) repeats 4 times. so 24/4 = 6 (Each digit will repeat 6 times in each place)

STEP 2: Sum of individual distinct digits (1,2,5,7) = 15

STEP 3: sum of individual distinct digits * No of times each digit got repeated at each place = 15*6 =90

STEP 4 : As the question is based on 4 digit integers 1111 * 90 = 99990

If it was 3 digit number calc 111 * 90 = 9990

Hope this helps you...

-- Shan

Same as the way i did it...

Hmmm great... I learnt this from my prep center... _________________

GMAT - Practice, Patience, Persistence Kudos if u like

Re: The addition problem above shows four of the 24 [#permalink]
26 Dec 2012, 23:42

vomhorizon wrote:

1,257 1,275 1,527 ........ ........ +7,521

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,5 and 7 exactly once in each integer. What is the sum of the 24 integers ?

Re: The addition problem above shows four of the 24 [#permalink]
20 Nov 2014, 17:24

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 [#permalink]
21 Nov 2014, 03:03

Expert's post

vomhorizon wrote:

1,257 1,275 1,527 ........ ........ +7,521

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,5 and 7 exactly once in each integer. What is the sum of the 24 integers ?

Re: The addition problem above shows four of the 24 [#permalink]
20 Apr 2015, 02:54

vomhorizon wrote:

1,257 1,275 1,527 ........ ........ +7,521

The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,5 and 7 exactly once in each integer. What is the sum of the 24 integers ?

If we write all the 4 digit numbers then each column will sum to 90. so unit digit of sum shld be 0 Nd tens digit shld be 9. only ans E is possible answer. _________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

gmatclubot

Re: The addition problem above shows four of the 24
[#permalink]
20 Apr 2015, 02:54

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I was checking my phone all day. I wasn’t sure when I would receive the admission decision from Tepper. I received an acceptance from Goizueta in the early morning...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...