Quantitative Aptitude

**Method 1: Cyclicity of Powers** The units digit of powers of 7 follows a cycle of length 4: - $7^1 = 7 \equiv 7 \pmod{10}$ - $7^2 = 49 \equiv 9 \pmod{10}$ - $7^3 = 343 \equiv 3 \pmod{10}$ - $7^4 = 2401 \equiv 1 \pmod{10}$ Since the exponent $84$ is divisible by $4$ ($84 = 4 \times 21$): $7^{84} = (7^4)^{21} \equiv 1^{21} \equiv 1 \pmod{10}$ Thus, the remainder when $7^{84}$ is divided by $10$ is $1$. **Method 2: Euler's Totient Theorem** $\phi(10) = 10 \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{5}\right) = 4$. Since $\gcd(7, 10) = 1$, by Euler's theorem: $7^{\phi(10)} = 7^4 \equiv 1 \pmod{10} \implies 7^{84} = (7^4)^{21} \equiv 1^{21} = 1$