How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated.

Solution (By Examveda Team)

Test of divisibility for 3:
The sum of the digits of any number that is divisible by 3 is divisible by 3
For instance, take the number 54372
Sum of its digits is 5 + 4 + 3 + 7 + 2 = 21
As 21 is divisible by 3, 54372 is also divisible by 3
There are six digits viz., 0, 1, 2, 3, 4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits.

The sum of all the six digits 0, 1, 2, 3, 4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits is divisible by 3

Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either 0 or 3 while forming the five digit numbers.

Case 1
If we do not use '0', then the remaining 5 digits can be arranged in:
5! ways = 120 numbers.

Case 2
If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'.

The first digit from the left can be any of the 4 digits 1, 2, 4 or 5
Then the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways.

So, there will be 4 × 4! numbers = 4 × 24 = 96 numbers.

Combining Case 1 and Case 2, there are a total of 120 + 96 = 216, 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.