Where is the error in my Conditional Statement (PHP)?

Question

The challenge is:

If n is odd, print "Weird".
If n is even and in the inclusive range of 2 to 5, print "Not Weird".
If n is even and in the inclusive range of 6 to 20, print "Weird".
If n is even and greater than 20, print "Not Weird".

And My Code is:

<?php
$N=5;

if($N%2==0 && $N>20){
    echo "Not Weird";
}else{
    echo "Weird";
}
?>

And the problem is:

When I rut it locally it's okay. But when I submit it to HackerRank then it fails their test when comes to $N=5; case. Do I have any problems in my conditional according to the challenge?

Maybe this can help you: http://stackoverflow.com/questions/7959247/php-test-if-number-is-odd-or-even — node_modules, May 09 '16 at 08:36
@JonStirling if $N= 5 then this condition work ($N%2==0 && $N>20) .??? this condition work on even no. only and $N>20 with $N%2==0 never work — ram singh, May 09 '16 at 08:44

score 1 · Accepted Answer · answered May 09 '16 at 08:34

1

<?php
$n=5;
if($n%2==1){
    echo 'Weird';
}else{
    if($n >= 2 && $n <= 5){
        echo 'Not Weird';
    }elseif($n > 5 && $n <= 20){
        echo 'Weird';
    }else{
        echo 'Not Weird';
    }
}
?>

answered May 09 '16 at 08:34

Niroj Adhikary

1,775
18
30

score 1 · Answer 2 · answered May 09 '16 at 08:51

I'm surprised that it works for you locally. It shouldn't. Your "Not weird" condition only works for even numbers which are greater then 20. You didn't include any condition that would cover unique cases (Weird for 6, 8, 10 etc).

One working solution would be:

<?php
if ($N%2 == 0 && ($N < 6 || $N > 20)) {
    echo "Not Weird";
} else {
    echo "Weird";
}

First condition displays Not Weird only when number is even, and is lower then 6, or greater then 20. For all other numbers - odd, and even from range 6 - 20 it displays Weird.

Where is the error in my Conditional Statement (PHP)?

2 Answers2