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Check if an Octal number is Even or Odd
• Last Updated : 23 Feb, 2021

Given an Octal number N, check whether it is even or odd.
Examples:

```Input: N = 7234
Output: Even

Input: N = 333333333
Output: Odd```

Naive Approach:

Time Complexity: O(N)
Efficient approach: Since Octal numbers contain digits from 0 to 7, therefore we can simply check if the last digit is either ‘0’, ‘2’, ‘4’ or ‘6’ . If it is, then the given Octal number will be Even, else Odd.
Below is the implementation of the above approach.

## C++

 `// C++ code to check if a Octal``// number is Even or Odd` `#include ``using` `namespace` `std;` `// Check if the number is odd or even``string even_or_odd(string N)``{``    ``int` `len = N.size();` `    ``// Check if the last digit``    ``// is either '0', '2', '4',``    ``// or '6'``    ``if` `(N[len - 1] == ``'0'``        ``|| N[len - 1] == ``'2'``        ``|| N[len - 1] == ``'4'``        ``|| N[len - 1] == ``'6'``)``        ``return` `(``"Even"``);``    ``else``        ``return` `(``"Odd"``);``}` `// Driver code``int` `main()``{``    ``string N = ``"735"``;` `    ``cout << even_or_odd(N);` `    ``return` `0;``}`

## Java

 `// Java code to check if a Octal``// number is Even or Odd``class` `GFG{`` ` `// Check if the number is odd or even``static` `String even_or_odd(String N)``{``    ``int` `len = N.length();`` ` `    ``// Check if the last digit``    ``// is either '0', '2', '4',``    ``// or '6'``    ``if` `(N.charAt(len - ``1``) == ``'0'``        ``|| N.charAt(len - ``1``) == ``'2'``        ``|| N.charAt(len - ``1``) == ``'4'``        ``|| N.charAt(len - ``1``) == ``'6'``)``        ``return` `(``"Even"``);``    ``else``        ``return` `(``"Odd"``);``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String N = ``"735"``;`` ` `    ``System.out.print(even_or_odd(N));``}``}` `// This code is contributed by Rajput-Ji`

## Python 3

 `# Python 3 code to check if a Octal``# number is Even or Odd` `# Check if the number is odd or even``def` `even_or_odd( N):``    ``l ``=` `len``(N);` `    ``# Check if the last digit``    ``# is either '0', '2', '4',``    ``# or '6'``    ``if` `(N[l ``-` `1``] ``=``=` `'0'``or` `N[l ``-` `1``] ``=``=` `'2'``or``        ``N[l ``-` `1``] ``=``=` `'4'` `or` `N[l ``-` `1``] ``=``=` `'6'``):``        ``return` `(``"Even"``)``    ``else``:``        ``return` `(``"Odd"``)` `# Driver code``N ``=` `"735"` `print``(even_or_odd(N))` `# This code is contributed by ANKITKUMAR34`

## C#

 `    ` `// C# code to check if a Octal``// number is Even or Odd``using` `System;` `public` `class` `GFG{``  ` `// Check if the number is odd or even``static` `String even_or_odd(String N)``{``    ``int` `len = N.Length;``  ` `    ``// Check if the last digit``    ``// is either '0', '2', '4',``    ``// or '6'``    ``if` `(N[len - 1] == ``'0'``        ``|| N[len - 1] == ``'2'``        ``|| N[len - 1] == ``'4'``        ``|| N[len - 1] == ``'6'``)``        ``return` `(``"Even"``);``    ``else``        ``return` `(``"Odd"``);``}``  ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``String N = ``"735"``;``  ` `    ``Console.Write(even_or_odd(N));``}``}` `// This code contributed by Princi Singh`

## Javascript

 ``
Output:
`Odd`

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