Question

find the two digit number which has square of the sum of it's digit equal to the number obtained by reversing it's digit​

Answer 1

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

Answer 2

$\underline{\underline{\Huge\mathfrak{Answer:}}}$

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