Talk:First perfect square in base n with n unique digits: Difference between revisions

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 val it : int = 1026753849
 </pre><br>which seems to save the expected two thirds time. This is a little trickier when an extra digit is required because you would have to check each extra digit separately (they would have different Digital Roots).<br>The sequence 1 4 9 7 7 9 4 1 9 1 4 9 7 7 9 4 1 9 1 4 can be produced as (n*n)%9 -> 1 4 0 7 7 0 4 1 0 1 4 0 7 7 0 4 1 0 1 4 for n = 1 to 10 with the usual 0 means 9.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 22:03, 27 May 2019 (UTC)
+: //after watering the garden <br>
+the digital root of the square of start value n ( pandigital:all digits of [0..Base-1] ) is for even numbers always Base-1.<BR>
+base 10: 0123456789 -> 09 18 27 36 45 -> 0 -> 9  <BR>
+For odd numbers Base % 2 ( casting out 9 aka Base-1 the Base % 2 is left over)<BR>
+base 9: 012345678 -> 08 17 26 35 4 -> 4 <BR>
+Now you only use such n , that there dgt-root**2 is the dgt-root of a pandigital number. Aha!<br>
+for base 9:  dgt root of the square [0..8] = 0  1  4  1  0  1  4  1  0 .. 1  4  1  0 <br>
+So you only use numbers with digit root  2 and 6<br><br>
+Now base 15: dgt-root of pandigital value ist 7<br>
+dgt root of the square:  0  1  4  9  2 11  8  7  8 11  2  9  4  1 .. 0<br>
+So one has only to check every 14.th value.<b>Bravo</b>!<br>