y & z - represent the digits of the numbers
A - the original number
B - the number with reversed digit
Set up the equation for the original number. To do so, let y represent the ten's digit and z the unit's digit.
`A = y(10) + z`
`A = 10y + z`
Also, set up the equation for the number with reversed digit.
`B = z(10) + y`
`B = 10z + y`
Since the digits are reversed, the z here becomes the ten's digit and y the unit's digit.
Then, add A and B.
`A + B = 121`
Susbtitute the equations of A and B.
`10y + z + 10z + y = 121`
`11y + 11z = 121`
Divide both sides by 11 to simplify.
`y + z = 11`
Note that y and z represents digits of a number which indicates that their values vary from 0 to 9.
So, assign values of x and y that will give a sum of 11. Then, check if it satisfy the given condition.
y=2 & x=9 ==> A=29 & B=92 ==> `29+92=121`
y=3 & x=8 ==> A=38 & B=83 ==> `38+83=121`
y=4 & x=7 ==> A=47 & B=74 ==> `47+74=121`
y=5 & x=6 ==> A=56 & B=65 ==> `56+65=121`
Also, take note that in the problem. it does not indicate if the orginal number A is less than B. So, if we have A greater than B and add the two numbers, their sum is still 121.
Hence, there are 8 possible values of the original number. These are 29, 38, 47, 56, 65, 74, 83 and 92.