How many six digit positive integers that are divisible by 3 can be formed using the digits 0,1,2,3,4,5 and 6 without any of the digits getting repeating?

Answer is 216. 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. read less