Question

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

• A. : 01
• B. : 21
• C. : 61
• D. : 71

Hint:

Use Euler’s theorem and reduce the exponent modulo \( \phi(100) = 40 \) to simplify power modulo 100.

Answer 1

The Correct Option is A

We are asked to find the last two digits of \( 7^{2008} \), i.e., we want to compute: \[ 7^{2008} \mod 100 \] Step 1: Use Euler’s Theorem: Euler's theorem tells us that: \[ a^{\phi(n)} \equiv 1 \pmod{n} \quad \text{if } \gcd(a, n) = 1 \] Here, \( a = 7 \), \( n = 100 \), and \( \gcd(7, 100) = 1 \). Now calculate \( \phi(100) \): \[ \phi(100) = \phi(2^2 \cdot 5^2) = 100 \left(1 - \frac{1}{2} \right) \left(1 - \frac{1}{5} \right) = 100 \cdot \frac{1}{2} \cdot \frac{4}{5} = 40 \] \[ \Rightarrow 7^{40} \equiv 1 \pmod{100} \] Step 2: Reduce the exponent modulo 40: \[ 2008 \mod 40 = 2008 - 40 \cdot 50 = 8 \] \[ \Rightarrow 7^{2008} \equiv 7^8 \mod 100 \] Step 3: Compute \( 7^8 \mod 100 \): \[ 7^2 = 49
7^4 = (7^2)^2 = 49^2 = 2401
\Rightarrow 7^4 \mod 100 = 2401 \mod 100 = 1 \] \[ \Rightarrow 7^8 = (7^4)^2 = 1^2 = 1 \Rightarrow \boxed{7^8 \equiv 1 \pmod{100}} \] Hence, the last two digits of \( 7^{2008} \) are: \[ \boxed{01} \] \fbox{Final Answer: (A): 01}