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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 19"

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== Problem ==
 
== Problem ==
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Let <math>N</math> be the positive integer <math>7777\ldots777</math>, a <math>313</math>-digit number where each digit is a <math>7</math>. Let <math>f(r)</math> be the leading digit of the <math>r{ }</math>th root of <math>N</math>. What is<cmath>f(2) + f(3) + f(4) + f(5)+ f(6)?</cmath><math>(\textbf{A})\: 8\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 11\qquad(\textbf{D}) \: 22\qquad(\textbf{E}) \: 29</math>
  
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== Solution ==
 +
For notation purposes, let <math>x</math> be the number <math>777 \ldots 777</math> with <math>313</math> digits, and let <math>B(n)</math> be the leading digit of <math>n</math>. As an example, <math>B(x) = 7</math>, because <math>x = 777 \ldots 777</math>, and the first digit of that is <math>7</math>.
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Notice that <cmath>B(\sqrt{\frac{n}{100}}) = B(\sqrt{n})</cmath> ​for all numbers <math>n \geq 100</math>; this is because <math>\sqrt{\frac{n}{100}} = \frac{\sqrt{n}}{10}</math>, and dividing by <math>10</math> does not affect the leading digit of a number. Similarly, <cmath>B(\sqrt[3]{\frac{n}{1000}}) = B(\sqrt[3]{n}).</cmath>
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In general, for positive integers <math>k</math> and real numbers <math>n > 10^{k}</math>, it is true that
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<cmath>B(\sqrt[k]{\frac{n}{10^{k}}}) = B(\sqrt[k]{n}).</cmath>
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Behind all this complex notation, all that we're really saying is that the first digit of something like <math>\sqrt[3]{123456789}</math> has the same first digit as <math>\sqrt[3]{123456.789}</math> and <math>\sqrt[3]{123.456789}</math>.
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The problem asks for <math>B(\sqrt[2]{x}} + B(\sqrt[3]{x}} + B(\sqrt[4]{x}} + B(\sqrt[5]{x}} + B(\sqrt[6]{x}}</math>. From our previous observation, we know that
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<cmath>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 .</cmath>
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Therefore, <math>B(\sqrt[2]{x}) = B(\sqrt[2]{7.777 \dots})</math>. We can evaluate <math>B(\sqrt[2]{7.777 \dots})</math>, the leading digit of <math>\sqrt[2]{7.777 \dots}</math>, to be <math>2</math>. Therefore, <math>f(2) = 2</math>.
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Similarly, we have 
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<cmath>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 .</cmath>
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Therefore, <math>B(\sqrt[3]{x}) = B(\sqrt[3]{7.777 \ldots})</math>. We know <math>B(\sqrt[3]{7.777 \ldots}) = 1</math>, so <math>f(3) = 1</math>.
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Similarly, <math>B(\sqrt[4]{x}) = B(\sqrt[4]{7.777 \ldots})</math> and <math>B(\sqrt[4]{7.777 \ldots}) = 1</math>, so <math>f(4) = 1</math>. We also have <math>B(\sqrt[5]{x}) = B(\sqrt[5]{777.777 \ldots})</math> and <math>B(\sqrt[5]{777.777 \ldots}) = 3</math>, so <math>f(5) = 3</math>. Finally, <math>B(\sqrt[6]{x}) = B(\sqrt[6]{7.777 \ldots})</math> and <math>B(\sqrt[4]{7.777 \ldots}) = 1</math>, so <math>f(6) = 1</math>.
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We have that <math>f(2)+f(3)+f(4)+f(5)+f(6) = 2+1+1+3+1 = \boxed{\textbf{(A) } 8}</math>.
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~ihatemath123
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=20|num-b=18}}
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=20|num-b=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:19, 23 November 2021

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

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 $B(\sqrt[2]{x}} + B(\sqrt[3]{x}} + B(\sqrt[4]{x}} + B(\sqrt[5]{x}} + B(\sqrt[6]{x}}$ (Error compiling LaTeX. Unknown error_msg). 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 $f(3) = 1$.

Similarly, $B(\sqrt[4]{x}) = B(\sqrt[4]{7.777 \ldots})$ 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

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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