Question

How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Answer 1

Suppose the original three-digit number is \(ABC\) where A, B, and C represent the hundreds, tens, and units digit, respectively.
Given that when the digits are reversed to form \(CBA\), the number increases by 198. 
This implies:
\(CBA = ABC + 198\)
We can represent the numbers as:
\(ABC = 100A + 10B + C\)
\(CBA = 100C + 10B + A\)
Substituting into the given equation:
\(100C + 10B + A = 100A + 10B + C + 198\)
Combining like terms:
\(99C - 99A = 198\)
Divide both sides by 99:
\(C - A = 2\)
Since A and C are single-digit integers, there are limited combinations that satisfy the above equation:

\((A, C) = (0, 2), (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9)\) 
However, the number must be greater than 100, So the first pair (0, 2) is not valid. So, there are 7 combinations that satisfy the condition. 
Now, B can be any digit from 0 to 9 for each combination. Therefore, the total number of three-digit numbers that meet the criteria is: \(7 \times 10 = 70\)
There are 70 such three-digit numbers.

Answer 2

Given, Three digit numbers increase by 198 when the three digits are arranged in the reverse order.
\(100c + 10b + a - 100a - 10b - c = 198\)
\(99c - 99a = 198\)
\(c - a = 2\)
So, the difference between the hundreds place digit and the units place digit is 2.
Now, \(a\) can range from 1 to 7, \(a\) cannot be 0 as the initial number has 3 digits and cannot be 8 or 9 as then \(c\) would not be a single-digit number. Thus, there can be 7 cases.
\(b\) can take the value of any digit from 0 to 9, as it does not affect the answer.
So, the possible combinations something like : 1 _ 3, 2 _ 4, 3 _ 5, 4 _ 6, 5 _ 7, 6 _ 8, 7 _ 9.
We have 10 numbers for each combination.
Hence, The total numbers =\(10 \times 7=70\).