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re: Compute Smart, Not Hard VIEW POST

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re: (I don't want to give away the game ;)...)

1988 is 32 * 82, and that's it.

#!/usr/bin/perl
use warnings;
use strict;

for my $born (1800 .. 2005) {
    my $prod = 1;
    $prod *= $_ for split //, $born;
    my $n = sqrt $prod;
    next unless $n == int $n;

    printf "%d %d\n", $born, $born + $n if $born + $n > 2005
}

Is there a way to hide the solution? PerlMonks have <readmore> and <spoiler>.

Metaproblem: restate the problem so that it has a determined answer.

An honors pupil Francesco was born in a year where the product of the digits of the year equals the square of some natural number, n. The year is now 2005, and he's waiting for the year when his age will be equal to that natural number, n. In how many years will Francesco celebrate the date?

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