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

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Question :

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

Started 10 months ago by Shashank in

0 285

Answer : The answer is '70'

Explanatory Answer

Let's consider a three-digit number to be ‘abc’.
Given that, 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.
The possible combinations are:
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 are 70.