BASH:check if a number is even or odd

In a bash script you might need to check if a var is even or odd ,

in my case im going to use a if to make that check

#!/bin/bash
for i in $(seq 10);do
    if (( $i % 2 )); then
        echo $i is odd
    else
        echo $i is even
    fi
done

You can use this simple code to make that check and display the result.

the result will be :


1 is odd
2 is even
3 is odd
4 is even
5 is odd
6 is even
7 is odd
8 is even
9 is odd
10 is even

5 thoughts on “BASH:check if a number is even or odd”

Tobi says:

6 December 2018 at 17 h 18 min

How would you input this script together to find a prime number?
Fabien says:

6 December 2018 at 17 h 37 min

Finding prime is way more complex to simply checking if the number is even or odd
a quick search gave me sthis : https://stackoverflow.com/questions/39002533/prime-number-script-checker
good luck
Fernando Basso says:

28 August 2019 at 10 h 32 min

You could also use a range instead of `seq`:

for i in {1..99}
CC says:

22 October 2020 at 18 h 21 min

Tested on Mac and Linux. None of them worked. The syntax is totally wrong. Why does this result appear on the top of google search???
DJM says:

4 December 2020 at 19 h 50 min

CC – The same thing happened to me when I cut-n-pasted the code. When I deleted the leading spaces on every line, it worked fine.