2021 Fall AMC 10B Problem 19

The number $N$ satisfies $$7\cdot10^{312}\lt N\lt 8\cdot10^{312}.$$ Multiplying or dividing by a power of $10$ only shifts the decimal point, so we only need the leading factor left after taking out the largest convenient power of $10.$ For $r=2,$ $\sqrt N$ has leading factor between $\sqrt7$ and $\sqrt8,$ so $f(2)=2.$ For $r=3,$ the leading factor is between $\sqrt[3]{7}$ and $\sqrt[3]{8},$ so $f(3)=1.$ For $r=4,$ the leading factor is between $\sqrt[4]{7}$ and $\sqrt[4]{8},$ so $f(4)=1.$ For $r=5,$ since $312=5\cdot62+2,$ the leading factor is between $\sqrt[5]{700}$ and $\sqrt[5]{800}.$ Since $3^5\lt700\lt800\lt4^5,$ $f(5)=3.$ For $r=6,$ the leading factor is between $\sqrt[6]{7}$ and $\sqrt[6]{8},$ so $f(6)=1.$ The sum is $2+1+1+3+1=8.$ Thus, the answer is A .