tonebeeze wrote:How many five-digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?
a. 15
b. 96
c. 120
d. 181
e. 216
OA = E
If the sum of the digits of integer N is a multiple of 3, then N itself is a multiple of 3.
Adding 5 of the digits above, there are 2 ways to get a sum that is a multiple of 3 if no digit is repeated:
1+2+3+4+5 = 15 and 0+1+2+4+5 = 12.
Number of ways to arrange 1,2,3,4,5 = 5! = 120.
Number of 5-digit integers composed of 0,1,2,4,5:
Ten-thousands digit can be 1,2,4,5 = 4 choices.
Number of ways to arrange the remaining 4 digits = 4! = 24.
Combining our choices for the digits, we get:
Number of possible integers = 4*24 = 96.
Thus, total possible integers = 120+96 = 216.
The correct answer is
E.
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