If the number 481*865 is completely divisible by 9, then the smallest whole…

Concept: a number is divisible by 9 exactly when the sum of its digits is divisible by 9. The reason is that 10 leaves a remainder of 1 when divided by 9, and so does every power of 10; a number therefore leaves the same remainder on division by 9 as the plain sum of its digits.

Application: the digit in the starred place is unknown, so write the digit sum with that digit as x and force the total to be a multiple of 9.

1. The six known digits are 4, 8, 1, 8, 6 and 5, and they add to 4 + 8 + 1 + 8 + 6 + 5 = 32.

2. Writing x for the digit in the starred place, the digit sum of 481x865 is 32 + x, where x is a single digit from 0 to 9.

3. Divisibility by 9 requires 32 + x to be a multiple of 9. Since 32 = 9 × 3 + 5, the known digits already carry a remainder of 5, so x must supply the missing 4.

4. x = 4 gives 32 + 4 = 36 = 9 × 4, a multiple of 9. The next value that would work, x = 13, is not a single digit, so 4 is the only admissible digit and hence the smallest.

Cross-check: every single-digit choice can be tested against the rule directly.

Substituting the digit gives 4814865, and 4814865 divided by 9 is exactly 534985 with no remainder. So the smallest whole number that can replace * is 4.