Bunuel wrote:
A five digits number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5, without repetition. What is the total number of ways this can be done?]
(A) 216
(B) 240
(C) 600
(D) 3125
(D) 3325
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions 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 0If 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