Self-referring

Number Theory Level 2

Find the two-digit number such that the square of the sum of its digits is equal to itself.

Can you prove it is the only integer with such property besides $0$ and $1$ ?

1 solution

Mr. India
Mar 23, 2019

Let number be $10a+b$

Given : $(a+b)^2=10a+b$

$a(a+2b)+b.b=10a+b×1$

Equating the coefficients of $a$ and $b$ ,

$a+2b=10,b=1$

So, $a+2(1)=10$

Or, $a=8$

Number is $\boxed{\boxed{81}}$

Note : this method doesn't work everytime