Understanding the Question:

• We are looking for a 3-digit number 𝐷 where the ratio of the number to the sum of its digits is least.
• We also need to find the difference between the digit at the hundred's place and the digit at the unit's place of 𝐷.

Approach

First, let's denote the 3-digit number as 𝐷, where  with A, B, and C representing the digits at the hundreds, tens, and units places, respectively.

Hence, D = 100A+ 10B + C

The sum of the digits of D

The ratio of 𝐷 to the sum of its digits is given by (assume it as r)

Demand of the question:

•   We are looking to minimize the ratio r.

•   Approach: For a minimum ratio (r), the numerator should be minimum and the denominator should be maximum.

•   i.e. we need to minimize D, and S should be is maximum.

Solution

First of all, we’ll try to maximize S:

If we take the maximum for all three, i.e. A = 9, B = 9, and C = 9.

Now, we’ll try to minimize D:

D will be minimized if A is minimum. i.e. A = 1 (as A ≠0), B = 9, and C = 9.

Just check out for any random middle number, e.g. 399, i.e. A = 3, B = 9, and C = 9.

The ratio gradually increases, with the ratio for 999 being 999/27 = 37.

So, the number with the least ratio is 199.

Hence The difference between the hundreds digit and the units digit = 9 - 1 = 8.

Therefore, option (c) is the correct answer.