SOLUTION: Find the 2 digit number which has the square of the sum of its digits equal to the number obtained by reversing it digit?
Algebra ->
Test
-> SOLUTION: Find the 2 digit number which has the square of the sum of its digits equal to the number obtained by reversing it digit?
Log On
You can put this solution on YOUR website! 10a+b = a two digit number
:
write what it says:
:
"A 2 digit number which has the square of the sum of its digits
equal to the number obtained by reversing it's digit?"
(a+b)^2 = 10b + a
We have two unknown with only one equation
Assume a = 1, find b
(1+b)^2 = 10b + 1
1 + 2b + b^2 = 10b + 1
:
b^2 + 2b - 10b = 1 - 1
b^2 - 8b = 0
divide both sides by b
b - 8= 0
b = 8
then the number is 18; (9^2 = 81)
:
I don't think there is any other number, except perhaps 10
(1+0)^2 = 0 + 1
