CAT 2017 Slot 2 — QA Question 30

For the 4 digit number to be divisible by 6, the sum of the digits of the 4 digit number has to be a multiple of 3 and the unit (right most) digit has to be an even number.
 
The combination of 4 digits that add upto a multiple of 3 are (0, 2, 3, 4), (0, 2, 4, 6) and (2, 3, 4, 6)

Case 1: Combination (0, 2, 3, 4)
For this combination the required number can start with 2, 3 or 4.
First digit 2: The last digit can be 0 or 4. The 2^nd and 3^rd digit from the left can be one 0/4 or 3. So a total of 4 possibilities.
First digit 4: Here too the number of possibilities is 4 as the 2^nd and 3^rd digit from the left can be one of 0/2 or 3.
First digit 3: Digits 0, 2 and 4 can be in any order as the 2^nd, 3^rd and 4^th digit from the left. So a total of 3! or 6 possibilities.
So number of possibilities with combination (0, 2, 3, 4) = 4 + 4 + 6 = 14

Case 2: Combination (0, 2, 4, 6)
The first digit can be anyone of 2, 4 or 6 i.e., 3 possibilities. The 2nd digit can be anyone from 0/2/4/6 but apart from the one used in the left most digit i.e., 3 possibilities. The 3^rd digit from the left will be any one from 0/2/4/6 but apart from the 2 digits used in the leftmost or 2^nd leftmost digit i.e., 2 possibilities. The extreme right digit will be the last digit after 3 out of these 4 digits are used in the first 3 digits from the left.
So total number of possibilities = 3 × 3 × 2 = 18

Case 3: Combination (2, 3, 4, 6)
The units digit has to be one amongst 2, 4 or 6 i.e., 3 possibilities. So first 3 digits from left will be 3 and two amongst 2/4/6 (i.e., except that digit which is not chosen as the right most digit) and can be arranged in 3! or 6 ways
So number of possibilities with combination (2, 3, 4, 6) is 6 × 3 = 18

So number of 4 digits numbers that can be formed using digits 0, 2, 3, 4 and 6 is 14 + 18 + 18 = 50

Hence, 50.