$(4^{61} + 4^{62} + 4^{63} + 4^{64})$ is divisible by

### Concept & Formula This problem tests the factorization of polynomials with common exponential bases. We can simplify the sum by factoring out the lowest power of the base, transforming addition into multiplication. ### Step-by-Step Solution * **Given:** The expression $(4^{61} + 4^{62} + 4^{63} + 4^{64})$. * **Factor out the smallest power:** Take out $4^{61}$ as a common multiplier. $$4^{61} (1 + 4^1 + 4^2 + 4^3)$$ * **Simplify the terms inside the parentheses:** $$4^{61} (1 + 4 + 16 + 64)$$ $$4^{61} (85)$$ * **Find the prime factors:** We now have $4^{61} \times 85$. Prime factorize 85 to find the divisors. $$85 = 5 \times 17$$ So, the expression is $4^{61} \times 5 \times 17$. * **Deduction:** Because 17 is one of the distinct prime factors of the resulting product, the original expression is entirely divisible by 17. ### Exam Strategy & Shortcut Skip writing out the powers if you know them mentally. Instantly write: $4^{61}(1+4+16+64) \rightarrow 4^{61}(85)$. Notice that 85 ends in 5, meaning it is divisible by 5. Mentally calculate $85 \div 5 = 17$. Scan the options for 17. ### Common Pitfall A frequent mistake is forgetting the $1$ when factoring $4^{61}$ out of itself, mistakenly writing $4^{61}(4 + 16 + 64) = 4^{61}(84)$, which would lead to incorrect divisor assumptions. ### Final Answer Therefore, the correct answer is **17**.