What is the remainder when $4^{61}$ is divided by 51?

### Concept & Logic This problem tests your ability to use **Modular Exponentiation** and cyclical patterns. To find the remainder of a massive power, we evaluate smaller powers of the base to find a number that leaves a remainder of $1$ or $-1$ when divided by the divisor. $$a^n \pmod k$$ ### Step-by-Step Solution * **Calculation:** Let's find a power of 4 that is close to a multiple of 51 by testing powers sequentially. * $4^1 = 4$ (Remainder 4) * $4^2 = 16$ (Remainder 16) * $4^3 = 64$ (Since $64 = 51 + 13$, Remainder is 13) * $4^4 = 256$ * Check 256 against multiples of 51. $51 \times 5 = 255$. * Therefore, $256 = 255 + 1$. * This means $4^4 \equiv 1 \pmod{51}$. This is our key cycle length! * Now, we express the original exponent (61) in terms of our cycle (4). $$61 = 4 \times 15 + 1$$ * Rewrite the original expression $4^{61}$ using exponent rules: $$4^{61} = (4^4)^{15} \times 4^1$$ * Substitute the remainder we found for $4^4$: $$(4^4)^{15} \times 4^1 \equiv (1)^{15} \times 4 \pmod{51}$$ $$\equiv 1 \times 4 \pmod{51}$$ * The final remainder is 4. * Comparing this to the given options (20, 41, 50), 4 is not listed. ### Exam Strategy & Shortcut Whenever dealing with "remainder of a huge power" problems, never try to calculate the actual number. Always run a quick mental cycle: $4^1, 4^2, 4^3, 4^4$. Look for proximity to multiples of the divisor (51, 102, 153, 204, 255...). Spotting that $4^4 (256)$ is exactly 1 above $51 \times 5 (255)$ instantly collapses the problem. You just divide the power 61 by 4, get a remainder of 1, and evaluate $4^1 = 4$. ### Common Pitfall Students often doubt their math when the correct numerical answer isn't explicitly listed in the options, causing them to re-do calculations and waste time rather than confidently selecting "None of these". Trust your cycle arithmetic. ### Final Answer Therefore, the correct answer is **None of these**.