Question 183641
Let {{{t}}}= the tens digit of the original number
Let {{{u}}}= the units digit of the original number
given:
The original number is {{{10t + u}}}
The number with the digits reversed is {{{10u + t}}}
(1) {{{10u + t = 10t + u + 9}}}
(2) {{{u + t = 11}}}
---------------------
From (1)
(1) {{{10u + t = 10t + u + 9}}}
(1) {{{10u - u + t - 10t = 9}}}
(1) {{{9u - 9t = 9}}}
(1) {{{u - t = 1}}}
Add (1) and (2)
(2) {{{u + t = 11}}}
(1) {{{u - t = 1}}}
(3) {{{2u = 12}}}
{{{u = 6}}}
And, since
(2) {{{u + t = 11}}}
(2) {{{6 + t = 11}}}
(2) {{{t = 5}}}
The original number is 56
check:
(1) {{{10u + t = 10t + u + 9}}}
(1) {{{60 + 5 = 50 + 6 + 9}}}
(1) {{{65 = 65}}}
(2) {{{u + t = 11}}}
(2) {{{6 + 5 = 11}}}
(2) {{{11 = 11}}}
OK