Wikipedia:Reference desk/Archives/Mathematics/2008 December 27

= December 27 =

Square numbers
I got a 202-digit square number:
 * 9....9 z 0....0 9

First there are 100 times a nine, then an unknown cipher called z, then 100 times a zero and then at last again a nine. So the 102nd number from right is not readable. How can you get this number? (Sry for my a bit broken English.) --85.178.8.133 (talk) 22:27, 27 December 2008 (UTC)


 * Take a look at 972, 9972, 99972, 999972 ... Spot the pattern. Prove it continues. (Hint: expand (10n-3)2). Gandalf61 (talk) 23:28, 27 December 2008 (UTC)
 * (and just to make sure that no other z would work, one may observe: in order to produce a 202-digit square, 10101 is too large; in order to get also the 100 initial nines, 10101- 6 is too small; and between these two the only number that produces the last nine when squared is the one written by Gandalf61. So the solution is unique even hiding all zeros...) PMajer (talk) 09:35, 28 December 2008 (UTC)