Question 147014
Let the first digit in the original number be A.
Let the second digit in the original number be B.
1.{{{A+B=12}}}
The number can be represented as
{{{10*A+B}}}
The number with digits transposed would be
{{{10*B+A}}}
Putting all those together would yield,
{{{10*B+A=10*A+B+54}}}
{{{-9A+9B=54}}}
2.{{{-A+B=6}}}
Add eq.1 and eq.2 and solve for B.
{{{A+B-A+B=12+6}}}
{{{2B=18}}}
{{{B=9}}}
From 1,
{{{A+B=12}}}
{{{A+9=12}}}
{{{A=3}}}
Original number is 39.
Transposed number is 93.