The total number of six digit numbers, formed using the digits 4, 5, 9 only and divisible by 6, is ____________.

Solution

Maths sir · 4 steps

Six means even and a multiple of 3: the last digit is forced to 4, and then the fifth digit is forced too.

1. Divisible by 2. A number divisible by 6 must be even. Of 4, 5 and 9, only 4 is even, so the units digit must be 4.
2. Divisible by 3. The digit sum must be a multiple of 3. The digits 4, 5, 9 leave remainders 1, 2, 0 when divided by 3. Each remainder appears exactly once.
3. Fill the places. Choose the first four digits freely: 3⁴ = 81 ways. Whatever their sum, exactly one of 4, 5, 9 in the fifth place makes the total, including the final 4, a multiple of 3. So the fifth digit has exactly 1 choice.

   3 × 3 × 3 × 3 × 1 × 1 = 81

4. So. There are 81 such six-digit numbers.

The common slip: 243. Fixing the units digit as 4 but forgetting the test for 3 lets all five other places vary freely, giving 3⁵ = 243, not 81.

The idea to keep. When the allowed digits cover every remainder mod 3 exactly once, fill all places but one freely. The last free place then has exactly one digit that fixes divisibility by 3.

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