Question

When the digits of a positive integer are written in reverse to form a new positive integer with the same number of digits(e.g., 1234 4321), the new number is 90 less than the original. What is the smallest possible value of the original number?

Answer 1

Case 1: a 2 digit number
in the original, let the unit digit be y and the tens digit be x
so the number is 10x + y and the reverse digit number is 10y + x
we have 10y + x - 10x - y = 90
9y - 9x = 90
y - x = 10 , no such numbers since both x and y must be less than or equal to 9
Case 2: a 3 digit number
arguing as above
100x + 10y + z - 100z - 10y - x = 90
99x - 99z = 90
11x - 11z = 10
11(x-y) = 10 none

case 3: 4 digit numbers
how about 1211 and 1121 , did not say we can't repeat digits.

1211 - 1121 = 90