Talk:Steady squares: Difference between revisions

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Revision as of 12:46, 21 December 2021

Reduce search range by observation of the results

By taking a look at the results,I tried to search for results by prepending a digit to n.
And there comes the solution:By prepending a digit to n, than n*n must end in n to be a steady square.
So only solutions of one digit before can be solutions. <lang pascal> function CalcSquare(n:LongInt;Pot10:byte); //pot10 is 10^(count of digits of n -1) //prepend one digit to n var

 dgt: LongInt;

begin

   //ex.: n= 5,Pot10 = 1, dgt= 2;-> n2= (2*10*1+5)^2 = 625
 For dgt := 1 to 9 do
 begin
   // n= 5 , dgt= 2*10 -> n2= 625
   n2 = sqr(dgt*10*Pot10+n);   
   n2 = dgt*dgt*100*Pot10*Pot10+2*dgt*10*pot10*n+n*n;

// term: dgt*dgt*100*Pot10*Pot10+2*dgt*10*pot10*n// always has ends in Pot10+1 "0" digits // so n*n must end in n, to be a steady square

 end;

</lang>

Note also that it must end in 1,5 or 6 (not 0 as any number ending in 0 squared ends with twice as many zeros).--Nigel Galloway (talk) 12:46, 21 December 2021 (UTC)

@@ Line 21: / Line 21: @@
   end;
 </lang>
+:Note also that it must end in 1,5 or 6 (not 0 as any number ending in 0 squared ends with twice as many zeros).--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 12:46, 21 December 2021 (UTC)