Question 456366
find the two digit number which has the square of the sum of its digits equal to the number obtained by reversing it digit?
:
Let x = the 10's digit
Let y = the units
then
10x+y = the number
:
(x + y)^2 = 10y + x
Rather than solving this equation, use some assumptions and logic
10y+x has to be perfect square which has two digits, not many of those.
Start with 81, then y=8, x=1
then 18 is the original number and sure enough (1+8)^2 = 81