Largest palindrome product: Difference between revisions

Find the largest palindrome made from the product of two n-digit numbers, where n ranges beyond 7,

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=={{header|ALGOL 68}}==

{{Trans|Wren}}

# returns n with the digits reversed #

PROC reverse = ( LONG INT v )LONG INT:

OD

OD

 

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{{Trans|Ring}}

Also showing the maximum for 2 and 4 .. 7 digit numbers. Tests for a better product before testing for palindromicity.

PROC is pal = ( LONG INT n )BOOL:

BEGIN

limit start *:= 10

OD

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<pre>

</pre>

 

 

=={{header|C#|CSharp}}==

===Main Task===

{{trans|Paper & Pencil}}

class Program {

 

Console.Write("{0} {1} μs", bs, sw.Elapsed.TotalMilliseconds * 1000.0);

}

{{out|Output @ Tio.run}}

<pre>913 x 993 = 906609 245.2 μs</pre>

 

===Stretch===

 

class Program {

if (isPal(p)) {

if (p > bp) bp = p;

p -= yy; c += 10; }

y -= ld == 7 ? 44 : 22; }

 

static void Main(string[] args) {

var sw = System.Diagnostics.Stopwatch.StartNew();

for (int i = 2, h = 1; i <= 10; h = ++i >> 1) {

a = new string('9', h) + new string('0', (h) - 1) + "1",

c = new string('9', h) + new string('0', i) + new string('9', h);

else doOne(i);

}

Console.Write("{0} sec", sw.Elapsed.TotalSeconds);

}

{{out|Output @ Tio.run}}

 

=={{header|F_Sharp|F#}}==

// Largest palindrome product. Nigel Galloway: November 3rd., 2021

let fN g=let rec fN g=[yield g%10; if g>=10 then yield! fN(g/10)] in let n=fN g in n=List.rev n

printfn "%d" ([for n in 100..999 do for g in n..999->n*g]|>List.filter fN|>List.max)

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<pre>

906609

</pre>

 

=={{header|Go}}==

{{trans|Wren}}

18 digit integers are within the range of Go's uint64 type though finding the result for 9-digit number products takes a while - around 15 seconds on my machine.

 

import "fmt"

}

}

 

{{out}}

Largest palindromic product of two 9-digit integers: 999980347 x 999920317 = 999900665566009999

</pre>

 

=={{header|Julia}}==

 

function twoprodpal(factorlength)

twoprodpal(i)

end

<pre>

For factor length 2, 91 * 99 = 9009

For factor length 11, 99999943851 * 99999996349 = 9999994020000204999999

For factor length 12, 999999000001 * 999999999999 = 999999000000000000999999

</pre>

 

=={{header|Paper & Pencil}}==

 

lowest possible 6 digit palindrome starting with 9 is 900009

979 x 931 = 911449‬, not a palindrome, so continue

979 x 921 = 901659‬, not a palindrome, and less than the best found so far, so stop

 

=={{header|Perl}}==

{{libheader|ntheory}}

use warnings;

use feature 'say';

say "Largest palindromic product of two @{[$l]}-digit integers: $f[1] × $f[0] = $p" and last LOOP;

}

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<pre>Largest palindromic product of two 2-digit integers: 91 × 99 = 9009

Largest palindromic product of two 6-digit integers: 999001 × 999999 = 999000000999

Largest palindromic product of two 7-digit integers: 9997647 × 9998017 = 99956644665999</pre>

 

=={{header|Raku}}==

 

my $p5 = Inline::Perl5.new();

$p5.use: 'ntheory';

}

$p

<pre>Largest palindromic product of two 2-digit integers: 91 × 99 = 9009

Largest palindromic product of two 3-digit integers: 913 × 993 = 906609

 

=={{header|Ring}}==

 

prod = 1

ok

end

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<pre>working...

Found in 6 iterations

done...</pre>

 

=={{header|Wren}}==

The approach here is to manufacture palindromic numbers of length 2n in decreasing order and then see if they're products of two n-digit numbers.

var r = 0

while (n > 0) {

if (nextN) break

}

 

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=={{header|XPL0}}==

int A, B;

[B:= 0;

];

IntOut(0, Max);

 

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