Let A be a 2n-digit number whose digits are all 1's and let B be a n-digit number whose digits are all 2's.Show that A-B is a perfect square.
A = sum(10^k,k=0 to 2n-1)
A=(10^(2n)-1)/9
B=2*sum(10^k,k=0 to n-1)
B=2*(10^n-1)/9
A-B=(10^(2n)-1)/9 - 2*(10^n-1)/9
A-B=(10^(2n)-2*10^n+1)/9
A-B=(10^n-1)^2/3^2
A-B=[(10^n-1)/3]^2
A-B=[3*(10^n-1)/9]^2
A-B=[3*sum(10^k,k=0 to n-1)]^2
A-B=[333....3]^2
thus the square root of A-B is the number consisting of n 3's. Thus A-B is a perfect square
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Posted by Daniel
on 2013-09-10 10:16:02 |
solution