D is a 3-digit number such that the ratio of the number to the sum of its digits is least. What is the difference between the digit at the hundred's place and the digit at the unit's place of D?

To find the least ratio of a 3-digit number to the sum of its digits, we need to make the number as small as possible while making the sum of its digits as large as possible. Let the 3-digit number be 100x plus 10y plus z. The sum of the digits is x plus y plus z. The ratio is minimized when the numerator is small and the denominator is large. To maximize the sum of digits, we should choose 9 for the tens and units places. To keep the value of the number small relative to the sum, the hundred's digit should be as small as possible but not zero. Testing numbers with large digit sums: For 199, the ratio is 199 divided by 19, which is approximately 10.47. For 100, the ratio is 100 divided by 1, which is 100. For 999, the ratio is 999 divided by 27, which is 37. The ratio decreases as the hundred's digit decreases and the other digits increase. The minimum ratio for a 3-digit number occurs at 199. In the number 199: The digit at the hundred's place is 1. The digit at the unit's place is 9. The difference between them is 9 minus 1, which equals 8. Therefore, the correct option is C.