How do you perform arithmetic in Bash so that 3 + 3 prints 6?

• A. $((3 + 3))
• B. $[3 + 3]
• C. expr 3 + 3
• D. let result=3+3; echo $result

Answer: (A)

Arithmetic expansion in Bash is performed with the syntax $(( ... )). The expression inside is evaluated as integer arithmetic by the shell, and the result replaces the expansion. So echo $((3 + 3)) asks Bash to compute 3 + 3 and substitute 6, which then gets printed. This method is built-in, fast, and works in Bash (and POSIX shells), and spaces around operators are optional, so you can also write echo $((3+3)). The other options rely on older or external mechanisms: the deprecated $[ ... ] form, or the external expr program, or the let builtin in a way that requires additional steps to print the result. The straightforward and idiomatic way to get 6 is the arithmetic expansion shown.