Question

What are the last two digits of \(7^{2008}\)?

• A. 21
• B. 61
• C. 01
• D. 41
• E. 81

Hint:

For finding last digits, use modular exponentiation and Euler’s theorem to reduce large powers quickly.

Answer 1

The Correct Option is C

We want \(7^{2008} \bmod 100\). Since \(\gcd(7,100) = 1\), Euler's theorem applies: \[ \phi(100) = 40, \quad 7^{40} \equiv 1 \ (\bmod\ 100) \] Now, \(2008 = 40 \times 50 + 8\), so: \[ 7^{2008} \equiv (7^{40})^{50} \times 7^8 \equiv 1^{50} \times 7^8 \ (\bmod\ 100) \] We compute \(7^4 = 2401 \equiv 1 \ (\bmod\ 100)\), hence \(7^8 \equiv 1\). Thus, the last two digits are \(\boxed{01}\).