How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?
Explanation:
Let the required three-digit number be ‘abc’.
According to the question:
‘abc’ + 198 = ‘cba’
⇒ 100a + 10b + c + 198 = 100c + 10b + a
⇒ 99c – 99a = 198
⇒ c – a = 2
∴ Value of (c, a) can be (9, 7) or (8, 6) or (7, 5) or (6, 4) or (5, 3) or (4, 2) or (3, 1) i.e., total 7 values.
For each of these 7 values of c and a, b can taken any of the 10 values from 0 to 9.
∴ Total such possible numbers = 7 × 10 = 70.
Hence, 70.
» Your doubt will be displayed only after approval.
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.