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

Correct Answer: 70

Let's analyze the given information:

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.