From d2d37f4a76f7866fec9eae447068dfe519faedb3 Mon Sep 17 00:00:00 2001
From: Steven Wilson <steven1170@zoho.eu>
Date: Fri, 14 Jan 2022 11:25:06 +0000
Subject: add solution week 147 task 2 in perl

---
 challenge-147/steven-wilson/perl/ch-02.pl | 40 +++++++++++++++++++++++++++++++
 1 file changed, 40 insertions(+)
 create mode 100644 challenge-147/steven-wilson/perl/ch-02.pl

diff --git a/challenge-147/steven-wilson/perl/ch-02.pl b/challenge-147/steven-wilson/perl/ch-02.pl
new file mode 100644
index 0000000000..65f41f0585
--- /dev/null
+++ b/challenge-147/steven-wilson/perl/ch-02.pl
@@ -0,0 +1,40 @@
+#!/usr/bin/env perl
+# Week 147 Task 2
+# Pentagon Numbers
+# Write a sript to find the first pair of Pentagon Numbers
+# whose sum and difference are also a Pentagon Number.
+# Answer: First pair is 210 and 330
+
+use strict;
+use warnings;
+use feature qw/ say /;
+
+my @pentagonal_numbers = qw/ 1 /;
+my $first_pair_found   = 0;
+my $number             = 2;
+
+while ( not $first_pair_found ) {
+    my $next_pn = pentagonal_number($number);
+    for (@pentagonal_numbers) {
+        if (    is_pentagonal_number( $next_pn - $_ )
+            and is_pentagonal_number( $next_pn + $_ ) )
+        {
+            say "First pair is $_ and $next_pn";
+            $first_pair_found = 1;
+            last;
+        }
+    }
+    push @pentagonal_numbers, $next_pn;
+    $number++;
+}
+
+sub pentagonal_number {
+    my $n = shift;
+    return ( ( 3 * ( $n * $n ) ) - $n ) / 2;
+}
+
+sub is_pentagonal_number {
+    my $x         = shift;
+    my $remainder = ( sqrt( 24 * $x + 1 ) + 1 ) % 6;
+    $remainder > 0 ? return 0 : return 1;
+}
-- 
cgit