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18 Mar 2020, 01:58
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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
18 Mar 2020, 03:02
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Bunuel
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
