Determine all possible digits a for which
9...9a0...09 is a square number. (Excluding the digit a there are 101 9's and 100 0's in the number)
Squares that look like this are of the form
(10^n - x)^2 = 10^(2n) - 2x*10^n + x^2
Since the number ends in ...09 we must have x=3.
The first two terms of the square give 10... (2n zeros) minus 60... (n zeros) which will makes the number begin with n 9's followed by a=4.
(Other endings x^2 give other values of a)
|
|
Posted by Jer
on 2020-11-12 08:33:12 |
Solution
