Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits 1,2,3,4,5 with repetition, is _________.

Solution

Maths sir · 4 steps

Divisible by 6 means even and a multiple of 3, so fix the last digit first, then sort digits by remainder.

1. Two conditions. A number is divisible by 6 when it is divisible by both 2 and 3. So the units digit must be 2 or 4, and the sum of the three digits must be a multiple of 3.
2. Group by remainder. Sort the digits by their remainder on division by 3. Remainder 0: only 3 (1 digit). Remainder 1: 1, 4 (2 digits). Remainder 2: 2, 5 (2 digits).
3. First two digits. Call the hundreds and tens digits a and b. If the units digit is 2, you need a + b to leave remainder 1. If it is 4, you need a + b to leave remainder 2. Multiply the group sizes for each remainder pair.

   Units digit	Remainder of a + b	Remainder pairs (a, b)	Count
   2	1	(0,1), (1,0), (2,2)	1 × 2 + 2 × 1 + 2 × 2 = 8
   4	2	(0,2), (2,0), (1,1)	1 × 2 + 2 × 1 + 2 × 2 = 8

4. So. Total = 8 + 8 = 16.

The common slip: 50. Checking only that the last digit is even gives 5 × 5 × 2 = 50. The digit sum must also be a multiple of 3.

The idea to keep. For a divisibility-by-3 condition with repetition allowed, sort the digits into remainder groups 0, 1, 2. Then count pairs of groups instead of listing every number.

Answer from the NTA final answer key. Solution by PrepWiser.