As written, the problem has no solution.
I must assume that there is a typo . . .
How many 3-digit numbers (in base 10, with no leading zero) when added to 495,
the digits of the sum is exactly reverse with the digit of the original number.
The original number is: .
Its reversal is: .
We are told: .
We have: .
. . . . . . . .
There are 4 choices: .
And can be any of the 10 available digits.
Therefore, there are 40 such numbers.