The total number of three-digit numbers, divisible by 3, which can be formed using the digits 1,3,5,8, if repetition of digits is allowed, is :

Solution

Maths sir · 3 steps

Replace each digit by its remainder mod 3, then count the patterns whose sum is a multiple of 3.

1. Remainders mod 3. A number is divisible by 3 exactly when its digit sum is. So only each digit's remainder on division by 3 matters. 3 leaves 0, 1 leaves 1, and both 5 and 8 leave 2.
2. Allowed patterns. Three remainders from {0, 1, 2} add up to a multiple of 3 only if all three are equal, or if they are 0, 1, 2 in some order. Repetition is allowed, so each place is filled independently.

   Remainder pattern	Digits used	Count
   0, 0, 0	only 3	1³ = 1 (333)
   1, 1, 1	only 1	1³ = 1 (111)
   2, 2, 2	5 or 8 in each place	2³ = 8
   0, 1, 2	3, 1, and one of 5, 8	2 × 3! = 12

3. So. The total is 1 + 1 + 8 + 12 = 22, which is option (b).

The idea to keep. For divisibility by 3, swap each digit for its remainder mod 3 and count the patterns that sum to a multiple of 3. Digits sharing a remainder multiply the count.

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