Largest palindrome product: Difference between revisions
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10 9999900001 x 9999999999 = 99999000000000099999
2.1622142 sec</pre>Wow! how did that go so fast? The results for the even-number-of-digit factors were manufactured by string manipulation instead of calculation (since the pattern was obvious). This algorithm can easily be adapted to BigIntegers for higher n-digit factors, but the execution time is unspectacular.
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls}}
<syntaxhighlight lang="Delphi">
function ReverseNum(N: int64): int64;
{Reverse the digit order of a number}
begin
Result:=0;
while N>0 do
begin
Result:=(Result*10)+(N mod 10);
N:=N div 10;
end;
end;
function IsPalindrome(N: int64): boolean;
{If the number is the same in }
{reverse order it is a palindrome}
var N1: int64;
begin
N1:=ReverseNum(N);
Result:=N = N1;
end;
procedure ShowPalindrome(Memo: TMemo; D,N1,N2: int64);
begin
Memo.Lines.Add(Format('%5D %5D %5D %10D',[D,N1,N2,N1 * N2]));
end;
procedure FindPalindromes(Digits: integer; var C1,C2: int64; Prog: TProgress);
{Find the largest palindrome derrived from two}
{ terms with the specified number of digits}
var I,J: cardinal;
var Prd,MinNum,MaxNum,Best: int64;
begin
Best:=0;
{Find the minimum and maximum values}
{ with the specified number of digits}
MinNum:=Trunc(Power(10,Digits-1));
MaxNum:=Trunc(Power(10,Digits))-1;
for I:=MinNum to MaxNum do
begin
{We can eliminate even factors and number ending in 5}
if ((I and 1)=0) or ((I mod 10)=5) then continue;
for J:=I+1 to MaxNum do
begin
{We can eliminate even factors and number ending in 5}
if ((J and 1)=0) or ((J mod 10)=5) then continue;
Prd:=I * J;
if not IsPalindrome(Prd) then continue;
if Assigned(Prog) then Prog(MulDiv(100,I-MinNum,MaxNum-MinNum));
{Save the largest palindromes}
if Prd>Best then
begin
Best:=Prd;
C1:=I; C2:=J;
end;
end;
end;
end;
procedure FindPalindromeMax(Memo: TMemo; Prog: TProgress);
var N1,N2: Int64;
var I: integer;
begin
Memo.Lines.Add('Digits F1 F2 Palindrome');
Memo.Lines.Add('-------------------------------');
for I:=2 to 4 do
begin
FindPalindromes(I,N1,N2,Prog);
ShowPalindrome(Memo,I,N1,N2);
end;
end;
</syntaxhighlight>
{{out}}
<pre>
Digits F1 F2 Palindrome
-------------------------------
2 91 99 9009
3 913 993 906609
4 9901 9999 99000099
</pre>
=={{header|F_Sharp|F#}}==
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