Let A be a 2ndigit number whose digits are all 1's and let B be a ndigit number whose digits are all 2's.Show that AB is a perfect square.
A = sum(10^k,k=0 to 2n1)
A=(10^(2n)1)/9
B=2*sum(10^k,k=0 to n1)
B=2*(10^n1)/9
AB=(10^(2n)1)/9  2*(10^n1)/9
AB=(10^(2n)2*10^n+1)/9
AB=(10^n1)^2/3^2
AB=[(10^n1)/3]^2
AB=[3*(10^n1)/9]^2
AB=[3*sum(10^k,k=0 to n1)]^2
AB=[333....3]^2
thus the square root of AB is the number consisting of n 3's. Thus AB is a perfect square

Posted by Daniel
on 20130910 10:16:02 