If the number 481*673 is completely divisible by 9, what is the smallest whole…

A number is completely divisible by 9 if and only if the sum of its digits is itself a multiple of 9. This is the standard digit-sum test for divisibility by 9.

1. Identify the known digits of 481*673: 4, 8, 1, *, 6, 7, 3.

2. Sum the known digits, excluding the unknown digit *: 4 + 8 + 1 + 6 + 7 + 3 = 29.

3. For the number to be divisible by 9, 29 plus the digit * must be a multiple of 9. Since * is a single digit from 0 to 9, the total sum ranges from 29 to 38.

4. The only multiple of 9 in the range 29 to 38 is 36.

5. So * = 36 - 29 = 7.

Substituting * = 7 gives the digits 4, 8, 1, 7, 6, 7, 3 with sum 4 + 8 + 1 + 7 + 6 + 7 + 3 = 36, and 36 / 9 = 4, confirming divisibility. No other digit from 0 to 9 works: 27 would need * = -2 (invalid) and 45 would need * = 16 (invalid). So 7 is the only valid digit, and hence also the smallest such whole number.