Find a five-digit number which has its digits reversed when it is multiplied by 9?

Question

When a two digit number is multiplied by sum of its digits, the product is 715 but when the reverse of a two digit number is multiplied by sum of its digits the product obtained is 99 less.
What is the sum of the digits of a number?

Answer 1

Original source: http://mathforum.org/library/drmath/view/54205.html

i found the answer on the above website
This struck my mind as soon as i read the post bcz its a kind of weird one cant forgot easily
so, here we go lets guess the 5 digit number is abcde and lets calculate the possibility

(9 here because we need to find the one which get reversed when multiplied by 9)
9*(10000*a+1000*b+100*c+10*d+e) = 10000*e+1000*d+100*c+10*b+a

        89999*a + 8990*b + 800*c = 910*d + 9991*e.

This implies that a = 1 and e = 9, by looking at last digits, as
before, and then

                  8990*b + 800*c = 910*d + 80

                    899*b + 80*c = 91*d + 8

This implies by looking at sizes that either b = 1 (so d = 1 and 899 +
80*c = 99, which has no solutions) or b = 0 (so 91*d - 80*c = 8, which
has the unique solution d = 8, c = 9). Thus there is just one 5-digit
solution 10989.