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Question
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?
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Solution
Five-digits numbers divisible by 3 are to be formed using the digits 0, 1, 2, 3, 4, and 5 without repetition.
For a number to be divisible by 3, the sum of its digits should be divisible by 3,
Consider the digits: 1, 2, 3, 4 and 5
Sum of the digits = 1 + 2 + 3 + 4 + 5 = 15
15 is divisible by 3.
∴ Any 5-digit number formed using the digits 1, 2, 1
Starting with the most significant digit, 5 digits are available for this place.
Since, repetition is not allowed, for the next significant place, 4 digits are available.
Similarly, all the places can be filled as:![]()
Number of 5-digit numbers
= 5 × 4 × 3 × 2 × 1 = 120
Now, consider the digits: 0, 1, 2, 4 and 5
Sum of the digits = 0 + 1 + 2 + 4 + 5 = 12 which is divisible by 3.
∴ Any 5-digit number formed using the digits
0, 1, 2, 4, and 5 will be divisible by 3.
Starting with the most significant digit, 4 digits are available for this place (since 0 cannot be used).
Since, repetition is not allowed, for the next significant place, 4 digits are available (since 0 can now be used).
Similarly, all the places can be filled as:![]()
Number of 5-digit numbers
= 4 × 4 × 3 × 2 × 1 = 96
Next, consider the digits: 0, 1, 2, 3, 4
Sum of the digits = 0 + 1 + 2 + 3 + 4 = 10 which is not divisible by 3.
∴ None of the 5-digit numbers formed using the digits 0, 1, 2, 3, and 4 will not be divisible by 3.
Further, no other selection of 5 digits (out of the given 6)
will give a 5-digit number, which is divisible by 3.
∴ Total number of 5adigit numbers divisible by 3
= 120 + 96 = 216
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