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18 Mar 2020, 01:58
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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62% (02:59) correct
38%
(03:02)
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based on 237
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Five digit numbers divisible by 9 are to be formed by using the digits 0, 1, 2, 3, 4, 7, 8 (without repetition). The total number of such numbers that can be formed is
A. 216
B. 214
C. 212
D. 200
E. 192
Are You Up For the Challenge: 700 Level Questions
A. 216
B. 214
C. 212
D. 200
E. 192
Are You Up For the Challenge: 700 Level Questions
18 Mar 2020, 02:57
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Minimum possible sum of 5 digits from 0, 1, 2, 3, 4, 7 and 8 = 0+1+2+3+4= 10
Maximum possible sum of 5 digits from 0, 1, 2, 3, 4, 7 and 8 = 2+3+4+7+8= 24
If Five digit numbers divisible by 9 are to be formed by using the digits 0, 1, 2, 3, 4, 7, 8 (without repetition), sum of digits must be 18.
Case 1- When 5 digits are 0,1,2, 7 and 8
Total possible numbers= 4*4*3*2*1= 96 (0 can't be at ten-thousands place)
Case 2- When 5 digits are 1,2, 3, 4 and 8
Total possible numbers= 5! = 120
Total numbers= 96+120= 216
Maximum possible sum of 5 digits from 0, 1, 2, 3, 4, 7 and 8 = 2+3+4+7+8= 24
If Five digit numbers divisible by 9 are to be formed by using the digits 0, 1, 2, 3, 4, 7, 8 (without repetition), sum of digits must be 18.
Case 1- When 5 digits are 0,1,2, 7 and 8
Total possible numbers= 4*4*3*2*1= 96 (0 can't be at ten-thousands place)
Case 2- When 5 digits are 1,2, 3, 4 and 8
Total possible numbers= 5! = 120
Total numbers= 96+120= 216
Bunuel
Five digit numbers divisible by 9 are to be formed by using the digits 0, 1, 2, 3, 4, 7, 8 (without repetition). The total number of such numbers that can be formed is
A. 216
B. 214
C. 212
D. 200
E. 192
Are You Up For the Challenge: 700 Level Questions
A. 216
B. 214
C. 212
D. 200
E. 192
Are You Up For the Challenge: 700 Level Questions
General Discussion
18 Mar 2020, 03:02
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Bookmarks
Bunuel
Five digit numbers divisible by 9 are to be formed by using the digits 0, 1, 2, 3, 4, 7, 8 (without repetition). The total number of such numbers that can be formed is
A. 216
B. 214
C. 212
D. 200
E. 192
A. 216
B. 214
C. 212
D. 200
E. 192
Divisibility Rule of 9: Sum of all digits should be divisible by 9
Possible set of 5 digits whose sum is divisible by 9 = {(1, 2, 3, 4, 8), (0, 1, 2, 7, 8)}
Number of 5 digit numbers possible using the digits {1,2,3,4,8} = 5! = 120
Number of 5 digit numbers possible using the digits {0, 1, 2, 7, 8} = 4*4*3*2*1 = 96 (ten thousand's place cannot be occupied by 0)
So, total 5 digit numbers possible = 120 + 96 = 216
Option A
