Q40. 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?

Let the 3-digit number be 100X + 10Y + Z. The sum of its digits is X + Y + Z. We want to minimize the ratio (100X + 10Y + Z) / (X + Y + Z). To minimize this ratio, we generally look for numbers where the hundreds digit (X) is small, and the sum of digits (X+Y+Z) is large.

Let's test numbers starting with the smallest possible hundreds digit, X=1: - For 100, ratio = 100 / (1+0+0) = 100. - For 109, ratio = 109 / (1+0+9) = 10.9. - For 119, ratio = 119 / (1+1+9) = 10.81.

- For 199, ratio = 199 / (1+9+9) = 199/19 ≈ 10.47. If we try numbers with X=2, e.g., 299, the ratio is 299 / (2+9+9) = 299/20 = 14.95, which is higher than 10.47. As the hundreds digit increases, the numerator grows much faster than the denominator, so the ratio will increase.

Thus, the number D that yields the least ratio is 199. For D=199, the digit at the hundred's place is 1, and the digit at the unit's place is 9. The difference between these digits is |1 - 9| = 8.

This question requires numerical reasoning and systematic testing, a common type in CSAT.