Solution:
We know that a number is divisible by 3 only when the sum of the digits is divisible by 3 . The given digits are 0,1,2,3,4,5
Here the possible number of combinations of 5 digits out of 6 are 5C4=5, which are as follows -
1+2+3+4+5=15=3×5( divisible by 3)
0+2+3+4+5=14 (not divisible by 3)
0+1+3+4+5=13 (not divisible by 3)
0+1+2+4+5=12=3×4( divisible by 3)
0+1+2+3+5=11 (not divisible by 3)
0+1+2+3+4=10 (not divisible by 3)
Thus the number should contain the digits 1,2,3,4,5 or the digits 0,1,2,4,5
Taking 1,2,3,4,5, the 5 digit numbers are =5!=120
Taking 0,1,2,4,5, the 5 digit numbers are =5!−4!=96
∴ Total number of numbers =120+96=216
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