Car in box riddle

#!/usr/local/bin/perl
# Author: Longree (2015-07-03)
# Computes answer to riddle: https://brilliant.org/problems/the-boxes/
# ( if my logic below is correct )

use warnings;
use strict;

# Box state matrix
# [box1, box2, box3] (1 if object in box)
my @boxstates = (
        [1, 0, 0],
        [0, 1, 0],
        [0, 0, 1]
);

# Test each box state
my $viablecount = 0;
for my $boxstate ( @boxstates )
{
        my $result = &runtest(@{$boxstate});
        printf("[%s] - %s\n", join(":", @{$boxstate}),
                ($result eq 1) ? "Viable" : "Not viable");
        $viablecount++ if ($result == 1);
}

# Output Result
if($viablecount < 1)
{
        print "None of the above :(\n";
}
elsif($viablecount == 1)
{
        print "One case exists! See above :)\n";
}
elsif(1 < $viablecount)
{
        print "Not enough information :(";
}

exit 0;

# Run test for given set of box states
sub runtest()
{
        my @boxstates = @_;

        my ($box1state, $box2state, $box3state) = @boxstates;

        my ($falsecount, $truecount) = (0, 0);

        &isinbox1($box1state) ? $truecount++ : $falsecount++;
        &notinbox2($box2state) ? $truecount++ : $falsecount++;
        &notinbox1($box1state) ? $truecount++ : $falsecount++;

        return (($falsecount == 2) && ($truecount == 1));
}

# statements 1
sub isinbox1()
{
        my $box1state = shift;

        return ($box1state == 1);
}

# statement 2
sub notinbox2()
{
        my $box2state = shift;

        return ($box2state == 0);
}

# statement 3
sub notinbox1()
{
        my $box1state = shift;

        return ($box1state == 0);
}