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2021 Fall AMC 10B Problems/Problem 19

Revision as of 13:43, 24 November 2021 by Ihatemath123 (talk | contribs)

Problem

Let $N$ be the positive integer $7777\ldots777$, 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\[f(2) + f(3) + f(4) + f(5)+ f(6)?\]$(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29$

Solution 1

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 \[B(\sqrt{\frac{n}{100}}) = B(\sqrt{n})\] ​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, \[B(\sqrt[3]{\frac{n}{1000}}) = B(\sqrt[3]{n}).\] In general, for positive integers $k$ and real numbers $n > 10^{k}$, it is true that \[B(\sqrt[k]{\frac{n}{10^{k}}}) = B(\sqrt[k]{n}).\] 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 \[B(\sqrt[2]{x}) = B(\sqrt[2]{\frac{x}{100}} = B(\sqrt[2]{\frac{x}{10,000}} = B(\sqrt[2]{\frac{x}{1,000,000}} = \ldots .\] Therefore, $B(\sqrt[2]{x}) = B(\sqrt[2]{7.777 \dots})$. We can evaluate $B(\sqrt[2]{7.777 \dots})$, the leading digit of $\sqrt[2]{7.777 \dots}$, to be $2$. Therefore, $f(2) = 2$.

Similarly, we have \[B(\sqrt[3]{x}) = B(\sqrt[3]{\frac{x}{1,000}} = B(\sqrt[3]{\frac{x}{1,000,000}} = B(\sqrt[3]{\frac{x}{1,000,000,000}} = \ldots .\] 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 \[B(\sqrt[5]{x}) = B(\sqrt[5]{777.777 \ldots})\] 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 $f(2)+f(3)+f(4)+f(5)+f(6) = 2+1+1+3+1 = \boxed{\textbf{(A) } 8}$.

~ihatemath123

Solution 2 (Condensed Solution 1)

Since $7777..7$ is a $313$ digit number and $\sqrt {7}$ is around $2.5$, we have $f(2)$ is $2$. $f(3)$ is the same story, so $f(3)$ is $1$. It is the same as $f(4)$ as well, so $f(4)$ is also $1$. However, $313$ is $3$ mod $5$, so we need to take the 5th root of $777$, which is between $3$ and $4$, and therefore, $f(5)$ is $3$. $f(6)$ is the same as $f(4)$, since it is $1$ more than a multiple of $6$. Therefore, we have $2+1+1+3+1$ which is $\boxed {(A) 8}$.

~Arcticturn

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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