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The addition problem above shows four of the 24 [#permalink]
20 Nov 2012, 23:53

3

This post was BOOKMARKED

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Difficulty:

25% (medium)

Question Stats:

78% (02:27) correct
22% (01:32) wrong based on 73 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

1

This post received KUDOS

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

1

This post was BOOKMARKED

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

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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

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

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