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vomhorizon
Joined: 03 Sep 2012
Last visit: 30 Mar 2018
Posts: 352
Given Kudos: 47
Location: United States
Concentration: Healthcare, Strategy
Schools: Duke '16 (A) McCombs '20 (A) Foster '20 (A)
GMAT 1: 730 Q48 V42
GPA: 3.88
WE:Medicine and Health (Healthcare/Pharmaceuticals)
Updated on: 21 Nov 2012, 04:30
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.
Last edited by vomhorizon on 21 Nov 2012, 04:30, edited 1 time in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:58) correct
30%
(02:00)
wrong
based on 442
sessions
History
Date
Time
Result
Not Attempted Yet
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
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
21 Nov 2012, 03:10
Kudos
Bookmarks
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.
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.
vomhorizon
Joined: 03 Sep 2012
Last visit: 30 Mar 2018
Posts: 352
Given Kudos: 47
Location: United States
Concentration: Healthcare, Strategy
Schools: Duke '16 (A) McCombs '20 (A) Foster '20 (A)
GMAT 1: 730 Q48 V42
GPA: 3.88
WE:Medicine and Health (Healthcare/Pharmaceuticals)
21 Nov 2012, 04:31
Kudos
Bookmarks
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..
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..
