An integer n, such as 1128, is called "sorted" if its digits are in sorted order. Find the largest integer n such that both n and n^{2} are sorted.
(In reply to
answer with proof by Daniel)
Let n = (2/3)*10^x + (1/3) = 6...67 [x1 6's and one 7]
then n^2 = [(2/3)*10^x]^2 + 2*(2/3)*10^x*(1/3)+ (1/3)^2
= (4/9)*10^(2x) + (4/9)*10^x + 1/9
= 4...44...4.444... + 4...4.444... + 0.111...
=4...48...89 [x 4's followed by x1 8's and one 9]

Posted by Jer
on 20130727 14:14:38 