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The addition problem above shows four of the 24 different integers
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Updated on: 21 Nov 2012, 04:30
Question Stats:
70% (01:29) correct 30% (01:14) wrong based on 172 sessions
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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 ? (A) 26,996 (B) 44,406 (C) 60,444 (D) 66,660 (E) 99,990
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Originally posted by vomhorizon on 21 Nov 2012, 00:53.
Last edited by vomhorizon on 21 Nov 2012, 04:30, edited 1 time in total.




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Re: The addition problem above shows four of the 24 different integers
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21 Nov 2012, 03:10
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 Adding up everything = 6666 + 13332 + 33330 + 46662 = 99990 Answer should be E. Kudos Please... If my post helped.
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Re: The addition problem above shows four of the 24 different integers
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21 Nov 2012, 04:31
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..
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Re: The addition problem above shows four of the 24 different integers
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21 Nov 2012, 06:36
Just always add up the thousand digit of the 24 different combinations, than you can see that A B C D are always smaller than 90k.



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Re: The addition problem above shows four of the 24 different integers
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21 Nov 2012, 07:38
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 ?
(A) 26,996 (B) 44,406 (C) 60,444 (D) 66,660 (E) 99,990 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
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Re: The addition problem above shows four of the 24 different integers
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27 Dec 2012, 00:42
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 ?
(A) 26,996 (B) 44,406 (C) 60,444 (D) 66,660 (E) 99,990 I used this formula... n!/n * sum of distinct digits * 111..1 = 4!/4 * 15 * 1111= 90 *1111 = 99990 For more explanation on that forumla: Sum of all Permutations of N Distince DigitsAnswer: E
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Re: The addition problem above shows four of the 24 different integers
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21 Nov 2014, 04:03
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 ?
(A) 26,996 (B) 44,406 (C) 60,444 (D) 66,660 (E) 99,990 Similar question to practice from OG: theadditionproblemaboveshowsfourofthe24differentin104166.html
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Re: The addition problem above shows four of the 24 different integers
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21 Nov 2014, 07:19
digits are 1,2,5 & 7. Total no =24 Taking unit digit sum=(1+2+5+7)*24/4=90
Taking tens digit sum=(1+2+5+7)*24/4=90
so last two digit of the sum of 24 nos will surely contains=90
Answer is E



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Re: The addition problem above shows four of the 24 different integers
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20 Apr 2015, 03:26
here we can use formula: (n1)!*sum of the digits*1111 so 3!*15*1111=99990



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Re: The addition problem above shows four of the 24 different integers
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20 Apr 2015, 03:54
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 ?
(A) 26,996 (B) 44,406 (C) 60,444 (D) 66,660 (E) 99,990 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.
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Re: The addition problem above shows four of the 24 different integers
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