2021 AMC 10 B Fall

Complete problem set with solutions and individual problem pages

Problem 19 Hard

Let N be the positive integer 7777 \ldots 777, a 313 -digit number where each digit is a 7 . Let f(r) be the leading digit of the r th root of N. What is f(2)+f(3)+f(4)+f(5)+f(6) ?(2021 AMC Fall 10B, Question #19)

  • A.

    8

  • B.

    9

  • C.

    11

  • D.

    22

  • E.

    29

Answer:A

Solution 1:

We can rewrite N as \frac{7}{9} \cdot 9999 \ldots, 999=\frac{7}{9} \cdot\left(10^{313}-1\right). When approximating values, as we will shortly do, the minus one will become negligible so we can ignore it. When we take the power of ten out of the square root, we'll be multiplying by another power of ten, so the leading digit will not change. Thus the leading digit of f(r) will be equal to the leading digit Then f(2) is the first digit of \sqrt{\frac{7}{9} \cdot(10)}=\sqrt{\frac{70}{9}}=\sqrt{7 \ldots .} \approx 2 f(3)-\sqrt[3]{\frac{7}{9} \cdot 10}=\sqrt[3]{\frac{70}{9}}=\sqrt[3]{7 \ldots .} \approx 1 f(4)-\sqrt[4]{\frac{7}{9} \cdot 10}=\sqrt[4]{\frac{70}{9}}=\sqrt[4]{7 \ldots} \approx 1 f(5)-\sqrt[5]{\frac{7}{9} \cdot 1000}=\sqrt[5]{\frac{7000}{9}}=\sqrt[6]{777 \ldots \ldots} \approx 3 . f(6)-\sqrt[6]{\frac{7}{9} \cdot 10}=\sqrt[6]{\frac{70}{9}}=\sqrt[6]{7 \ldots} \approx 1 The final answer is therefore 2+1+1+3+1=8=A

Solution 2:

For notation purposes, let x be the number 777 \ldots 777 with 313 digits, and let B(n) be the leading digit of n. As an example, B(x)=7, because x=777 \ldots 777, and the first digit of that is 7 . Notice that B\left(\sqrt{\frac{n}{100}}\right)=B(\sqrt{n}) for all numbers n \geq 100; this is because \sqrt{\frac{n}{100}}=\frac{\sqrt{n}}{10}, and dividing by 10 does not affect the leading digit of a number. Similarly, B\left(\sqrt[3]{\frac{n}{1000}}\right)=B(\sqrt[3]{n}) . In general, for positive integers k and real numbers n>10^{k}, it is true that B\left(\sqrt[k]{\frac{n}{10^{k}}}\right)=B(\sqrt[k]{n}) Behind all this complex notation, all that we're really saying is that the first digit of something like \sqrt[3]{123456789} has the same first digit as \sqrt[3]{123456.789} and \sqrt[3]{123.456789}. The problem asks for B(\sqrt[2]{x})+B(\sqrt[3]{x})+B(\sqrt[4]{x})+B(\sqrt[5]{x})+B(\sqrt[6]{x}) . From our previous observation, we know that B(\sqrt[2]{x})=B\left(\sqrt[2]{\frac{x}{100}}=B\left(\sqrt[2]{\frac{x}{10,000}}=B\left(\sqrt[2]{\frac{x}{1,000,000}}=\ldots\right.\right.\right. Therefore, B(\sqrt[2]{x})=B(\sqrt[2]{7.777 \ldots}). We can evaluate B(\sqrt[2]{7.777 \ldots}), the leading digit of \sqrt[2]{7.777 \ldots}, to be 2 . Therefore, f(2)=2. Similarly, we have B(\sqrt[3]{x})=B\left(\sqrt[3]{\frac{x}{1,000}}=B\left(\sqrt[3]{\frac{x}{1,000,000}}=B\left(\sqrt[3]{\frac{x}{1,000,000,000}}=\ldots\right.\right.\right. Therefore, B(\sqrt[3]{x})=B(\sqrt[3]{7.777 \ldots}). We know B(\sqrt[3]{7.777 \ldots})=1, so f(3)=1.

Next, B(\sqrt[4]{x})=B(\sqrt[4]{7.777 \ldots}) and B(\sqrt[4]{7.777 \ldots})=1, so f(4)=1. We also have B(\sqrt[5]{x})=B(\sqrt[3]{777.777 \ldots}) and B(\sqrt[5]{777.777 \ldots})=3, so f(5)=3. Finally, B(\sqrt[6]{x})=B(\sqrt[6]{7.777 \ldots}) and B(\sqrt[4]{7.777 \ldots})=1, so f(6)=1. We have that f(2)+f(3)+f(4)+f(5)+f(6)=2+1+1+3+1= (A) 8 .