A five digits number divisible by 3 is to be formed using the numbers

A number is divisible by 3 if sum of its digits is divisible by 3. From the given numbers we can form numbers that are divisible by 3 in 2 ways:
1) From the numbers 1,2,3,4, and 5: 5!=120 different numbers can be formed without repetition.
2) From the numbers 0,1,2,4 and 5: 5!-4!=120-24=96 (we subtract 4! from 5! because 0 can not be the first digit) different numbers can be formed without repetition.
In total 120+96=216 different numbers that are divisible by 3 can be formed from the given numbers without repetition.
IMO Answer is A)

Solution:

Recall that if the sum of the digits of a number is divisible by 3, then it is divisible by 3. We see that the sum of digits 0 to 5, inclusive, is 15, which is divisible by 3. However, there are 6 digits; so if we need to choose 5 of them and still want the sum of the digits to be divisible by 3, we must exclude a digit that is divisible by 3. That is, we must exclude either 0 or 3.

Case 1: Excluding 0

If we exclude 0, we will be using the digits 1 to 5, inclusive, to form a 5-digit number. Notice that the sum of these digits is 15, so no matter how we arrange these digits to form a 5-digit number, it will be divisible by 3. Therefore, there are 5! = 120 such numbers in this case.

Case 2: Excluding 3

If we exclude 3, we will be using the digits 0, 1, 2, 4, and 5 to form a 5-digit number. Notice that the sum of these digits is 12, so no matter how we arrange these digits to form a 5-digit number (assuming the digit 0 is allowed to be the first digit of the number), it will be divisible by 3. Therefore, there will also be 5! = 120 such numbers in this case if we permit the digit 0 as the first digit of the number. However, since we can’t have the digit 0 as the first digit of a number, we have to exclude one-fifth, or 24, of the 120 numbers. Therefore, the actual number of numbers in this case is 120 - 24 = 96.

Therefore, there are a total of 120 + 96 = 216 such numbers.

Answer: A

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