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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 19"
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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>.
 
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>.
  
−
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
+
The problem asks for
 +
<cmath>B(\sqrt[2]{x}) + B(\sqrt[3]{x}) + B(\sqrt[4]{x}) + B(\sqrt[5]{x}) + B(\sqrt[6]{x}).</cmath>
 +
 
 +
From our previous observation, we know that
 
<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>  
 
<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>  
 
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>.
 
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>.
Line 18: Line 21:
 
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>.
 
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>.
  
−
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>.
+
Next,
 +
<cmath>B(\sqrt[4]{x}) = B(\sqrt[4]{7.777 \ldots})</cmath>
 +
and <math>B(\sqrt[4]{7.777 \ldots}) = 1</math>, so <math>f(4) = 1</math>.
 +
 
 +
We also have
 +
<cmath>B(\sqrt[5]{x}) = B(\sqrt[5]{777.777 \ldots})</cmath>
 +
and <math>B(\sqrt[5]{777.777 \ldots}) = 3</math>, so <math>f(5) = 3</math>.
 +
 
 +
Finally,
 +
<cmath>B(\sqrt[6]{x}) = B(\sqrt[6]{7.777 \ldots})</cmath>
 +
and <math>B(\sqrt[4]{7.777 \ldots}) = 1</math>, so <math>f(6) = 1</math>.
  
 
We have that <math>f(2)+f(3)+f(4)+f(5)+f(6) = 2+1+1+3+1 = \boxed{\textbf{(A) } 8}</math>.
 
We have that <math>f(2)+f(3)+f(4)+f(5)+f(6) = 2+1+1+3+1 = \boxed{\textbf{(A) } 8}</math>.

Revision as of 11:22, 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 \[B(\sqrt[2]{x}) + B(\sqrt[3]{x}) + B(\sqrt[4]{x}) + B(\sqrt[5]{x}) + B(\sqrt[6]{x}).\]

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

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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