Question

Find the remainder when \( 2^{1040} \) is divided by 131.

• A. 1
• B. 3
• C. 5
• D. 7

Hint:

Apply Fermat’s Little Theorem when base and modulus are coprime and modulus is prime.

Answer 1

The Correct Option is A

Use Fermat’s Little Theorem: If \( p \) is prime and \( a \not\equiv 0 \pmod{p} \), then \[ a^{p-1} \equiv 1 \pmod{p} \] Here, \( a = 2 \), \( p = 131 \Rightarrow 2^{130} \equiv 1 \pmod{131} \)
Now write: \[ 1040 = 130 \times 8 \Rightarrow 2^{1040} = (2^{130})^8 \equiv 1^8 \equiv \boxed{1} \pmod{131} \]