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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 19"
Line 41: Line 41:
  
 
~Arcticturn
 
~Arcticturn
 +
 +
== Solution 3 ==
 +
First, we compute <math>f \left( 2 \right)</math>.
 +
 +
Because <math>N > 4 \cdot 10^{312}</math>, <math>\sqrt{N} > 2 \cdot 10^{166}</math>.
 +
Because <math>N < 9 \cdot 10^{312}</math>, <math>\sqrt{N} < 3 \cdot 10^{166}</math>.
 +
 +
Therefore, <math>f \left( 2 \right) = 2</math>.
 +
 +
Second, we compute <math>f \left( 3 \right)</math>.
 +
 +
Because <math>N > 1 \cdot 10^{312}</math>, <math>\sqrt[3]{N} > 1 \cdot 10^{104}</math>.
 +
Because <math>N < 8 \cdot 10^{312}</math>, <math>\sqrt[3]{N} < 2 \cdot 10^{104}</math>.
 +
 +
Therefore, <math>f \left( 3 \right) = 1</math>.
 +
 +
Third, we compute <math>f \left( 4 \right)</math>.
 +
 +
Because <math>N > 1 \cdot 10^{312}</math>, <math>\sqrt[4]{N} > 1 \cdot 10^{78}</math>.
 +
Because <math>N < 16 \cdot 10^{312}</math>, <math>\sqrt[4]{N} < 2 \cdot 10^{78}</math>.
 +
 +
Therefore, <math>f \left( 4 \right) = 1</math>.
 +
 +
 +
Fourth, we compute <math>f \left( 5 \right)</math>.
 +
 +
Because <math>N > 3^5 \cdot 10^{310}</math>, <math>\sqrt[5]{N} > 3 \cdot 10^{62}</math>.
 +
Because <math>N < 4^5 \cdot 10^{310}</math>, <math>\sqrt[5]{N} < 4 \cdot 10^{62}</math>.
 +
 +
Therefore, <math>f \left( 5 \right) = 3</math>.
 +
 +
Fifth, we compute <math>f \left( 6 \right)</math>.
 +
 +
Because <math>N > 1 \cdot 10^{312}</math>, <math>\sqrt[6]{N} > 1 \cdot 10^{52}</math>.
 +
Because <math>N < 2^6 \cdot 10^{312}</math>, <math>\sqrt[6]{N} < 2 \cdot 10^{52}</math>.
 +
 +
Therefore, <math>f \left( 6 \right) = 1</math>.
 +
 +
Therefore,
 +
<cmath>
 +
\begin{align*}
 +
f \left( 2 \right) + f \left( 3 \right) + f \left( 4 \right) + f \left( 5 \right) + f \left( 6 \right)
 +
& = 2 + 1 + 1 + 3 + 1 \\
 +
& = 8 .
 +
\end{align*}
 +
</cmath>
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(A) }8}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
 +
  
 
==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 22:57, 25 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 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

Solution 3

First, we compute $f \left( 2 \right)$.

Because $N > 4 \cdot 10^{312}$, $\sqrt{N} > 2 \cdot 10^{166}$. Because $N < 9 \cdot 10^{312}$, $\sqrt{N} < 3 \cdot 10^{166}$.

Therefore, $f \left( 2 \right) = 2$.

Second, we compute $f \left( 3 \right)$.

Because $N > 1 \cdot 10^{312}$, $\sqrt[3]{N} > 1 \cdot 10^{104}$. Because $N < 8 \cdot 10^{312}$, $\sqrt[3]{N} < 2 \cdot 10^{104}$.

Therefore, $f \left( 3 \right) = 1$.

Third, we compute $f \left( 4 \right)$.

Because $N > 1 \cdot 10^{312}$, $\sqrt[4]{N} > 1 \cdot 10^{78}$. Because $N < 16 \cdot 10^{312}$, $\sqrt[4]{N} < 2 \cdot 10^{78}$.

Therefore, $f \left( 4 \right) = 1$.


Fourth, we compute $f \left( 5 \right)$.

Because $N > 3^5 \cdot 10^{310}$, $\sqrt[5]{N} > 3 \cdot 10^{62}$. Because $N < 4^5 \cdot 10^{310}$, $\sqrt[5]{N} < 4 \cdot 10^{62}$.

Therefore, $f \left( 5 \right) = 3$.

Fifth, we compute $f \left( 6 \right)$.

Because $N > 1 \cdot 10^{312}$, $\sqrt[6]{N} > 1 \cdot 10^{52}$. Because $N < 2^6 \cdot 10^{312}$, $\sqrt[6]{N} < 2 \cdot 10^{52}$.

Therefore, $f \left( 6 \right) = 1$.

Therefore, \begin{align*} f \left( 2 \right) + f \left( 3 \right) + f \left( 4 \right) + f \left( 5 \right) + f \left( 6 \right) & = 2 + 1 + 1 + 3 + 1 \\ & = 8 . \end{align*}

Therefore, the answer is $\boxed{\textbf{(A) }8}$.

~Steven Chen (www.professorchenedu.com)


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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