Question

The last digit of $7^{100}$ is: }

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

7 has a cyclicity of 4. $7^4$, $7^8$, $7^{12}$... all end in 1.

Answer 1

The Correct Option is B

Step 1: Concept
Cyclicity of unit digits.

Step 2: Analysis
Powers of 7 follow a cycle of 4:
$7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401$. Cycle is $\{7, 9, 3, 1\}$.

Step 3: Calculation
Divide the power by 4: $100 \div 4$ gives remainder 0.
A remainder of 0 corresponds to the 4th position in the cycle.

Step 4: Conclusion
The last digit is 1. Final Answer: (B)