Question 465888
I think I've already solved this problem; there is no such positive integer. This is because the digit sum remains the same, so the remainder upon dividing by 9 will be the same.


To see this, take some random numbers like 583, 79, and 1056. Reversing their digits we get 385, 97, and 6501. If we take their differences we get


583 - 385 = 198 = 22*9
79 - 97 = -18 = -2*9
1056 - 6501 = -5445 = -605*9


In all these cases we are essentially adding or subtracting multiples of 9, and since 12 is not a multiple of 9 we cannot make a number increase by 12.


However this is entirely possible in other bases. If we have *[tex 12_{13}] (12 in base 13 = 15) and we reverse the digits we get *[tex 21_{13}], and


*[tex \LARGE 21_{13} - 12_{13} = 27_{10} - 15_{10} = 12_{10}]


See if you can find the smallest base-10 number that, when written in a different base, the property holds (here the base-10 number was 15).