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.