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 repeating?

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 are 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.                                                                                                                           

              FOR A NUMBER TO BE DIVISIBLE BY 3 THE SUM OF ITS DIGITS SHOULD BE DIVISIBLE BY 3 . SO, AMONG THE 6 DIGITS , 5 DIGITS

              WHOSE SUM IS DIVISIBLE BY 3 MUST BE SELECTED.

                           0+1+2+3+4+5 = 15 ( a multiple of 3)

               SO THE FIVE DIGITS CAN BE OBTAINED BY ELIMINATING A 3 MULTIPLE . 

                   15 - 3q = 3(5 - q) . SO IT IS DIVISIBLE BY 3.

                    AMONG 0,1,2,3,4,5 ;  0 AND 3 ARE THE MULTIPLES OF 3 .

                   SO, THE 5 DIGITS MUST BE  (1,2,3,4,5) OR (0,1,2,4,5)

                     FIRST LET US TAKE (1,2,3,4,5) . ACCORDING TO PERMUTATION LAW , IT CAN BE ARRANGED IN 5_p5 WAYS

                                                                                         5_P5  = 5 x 4 x 3 x 2 x 1 = 120

                                         (0,1,2,4,5) . IF ZERO IS THE FIRST DIGIT IT BECOMES A 4 DIGIT NUMBER .      

                   NUMBER OF  NUMBERS HAVING 0 AS FIRST DIGIT ARE                             

                                                  4 x 3 x 2 x 1 = 24                                 

                          SO, NUMBER OF 5 DIGIT NUMBER FORMED FROM (0,1,2,4,5)  ARE 120 - 24 = 96.

                             TOTAL  NUMBER OF 5 DIGIT NUMBER FORMED FROM  (0, 1,2,3,4,5)   ARE 120 + 96 = 216.