2021 AMC Fall 10B Problem 19

Since $N$ has $313$ digits, $N = 7.777\ldots \times 10^{312}$. To take an $r$th root cleanly, shift the decimal point so the power of $10$ is a multiple of $r$, then find which integer $d$ satisfies $d^r \le (\text{leading factor}) < (d+1)^r$.

• $r = 2$: $312 = 2 \cdot 156$, and $2^2 = 4 \le 7.77 < 9 = 3^2$, so $f(2) = 2$.
• $r = 3$: $312 = 3 \cdot 104$, and $1 \le 7.77 < 8 = 2^3$, so $f(3) = 1$.
• $r = 4$: $312 = 4 \cdot 78$, and $1 \le 7.77 < 16 = 2^4$, so $f(4) = 1$.
• $r = 5$: $312$ is not a multiple of $5$; write $N = 777.7\ldots \times 10^{310}$ with $310 = 5 \cdot 62$. Since $3^5 = 243 \le 777.7 < 1024 = 4^5$, $f(5) = 3$.
• $r = 6$: $312 = 6 \cdot 52$, and $1 \le 7.77 < 64 = 2^6$, so $f(6) = 1$.

The sum is $2 + 1 + 1 + 3 + 1 = 8$.

The answer is $\boxed{\textbf{(A)}\ 8}$.

The leading digit of $\sqrt[r]{N}$ depends only on $N$'s size relative to powers of $10$: once the exponent of $10$ is divisible by $r$, the root is (leading factor)$^{1/r}$ times a power of $10$, and the leading factor lies between $1$ and $1000$. This is scientific-notation reasoning; no calculator-level precision is needed.

• For the fifth root, dividing 312 by 5 and taking the leading digit of the fifth root of 7.77, which ignores that 312 is not a multiple of 5.
• Miscounting the exponent as $313$ instead of $312$ (a $313$-digit number is between $10^{312}$ and $10^{313}$), which shifts several answers.