How many 5 digit numbers can be formed which are divisible

First step:

We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.

We have six digits: 0, 1, 2, 3, 4, 5. Their sum=15.

For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.

Second step:

We have two set of numbers:
1, 2, 3, 4, 5 and 0, 1, 2, 4, 5. How many 5 digit numbers can be formed using these two sets:

1, 2, 3, 4, 5 --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.

0, 1, 2, 4, 5 --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, desired # would be total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=4!(5-1)=4!*4=96

120+96=216

Answer: A.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-five-digit-number-divisible-by-3-is-to-be-formed-using-136900.html