Getting even number using grep in bash [on hold]

Question

I have a range of numbers example 1 to 10 or 1 to 100. I used awk command and perl command to implement that. I want to use grep to get all even numbers and show it into terminal all even numbers.

I have a file named File.txt and contains the range of numbers: 1 2 3 4 5 6 7 8 9 0.

Using this code awk '. ($0%2)' File.txt awk '. ($0%2)' File.txt the output is 2 4 6 8 0.

Also with this code perl -ne 'print if($.%2 == 0)' File.txt the output is 2 4 6 8 0.

How can I use grep with the same output as perl and awk?

Answer 1

Given that even numbers are sequence of digits NOT ending with 02468 (and odd numbers ends with 13579), you can use:

   # Even
grep '^[0-9]*[02468]$'
   #  Odd
grep '^[0-9]*[13579]$'