Difference between revisions of "2021 Fall AMC 10B Problems/Problem 19"
(Created page with "== Problem == Three identical square sheets of paper each with side length <math>6{ }</math> are stacked on top of each other. The middle sheet is rotated clockwise <math>30^...") |
m (→Solution 1) |
||
(22 intermediate revisions by 12 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
+ | 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> | ||
− | + | ==Solution 1== | |
− | + | We can rewrite <math>N</math> as <math>\frac{7}{9}\cdot 9999\ldots999 = \frac{7}{9}\cdot(10^{313}-1)</math>. | |
+ | 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 <math>f(r)</math> will be equal to the leading digit of <math>\sqrt[r]{\frac{7}{9} \cdot 10^{313(\mod r)}}</math>. | ||
− | <math>(\textbf{A})\ | + | Then <math>f(2)</math> is the first digit of <math>\sqrt{\frac{7}{9}\cdot(10)} = \sqrt{\frac{70}{9}} = \sqrt{7.\ldots} \approx 2</math> |
+ | |||
+ | <math>f(3) - \sqrt[3]{\frac{7}{9} \cdot 10} = \sqrt[3]{\frac{70}{9}} = \sqrt[3]{7.\ldots} \approx 1</math>. | ||
+ | |||
+ | <math>f(4) - \sqrt[4]{\frac{7}{9} \cdot 10} = \sqrt[4]{\frac{70}{9}} = \sqrt[4]{7.\ldots} \approx 1</math>. | ||
+ | |||
+ | <math>f(5) - \sqrt[5]{\frac{7}{9} \cdot 1000} = \sqrt[5]{\frac{7000}{9}} = \sqrt[5]{777.\ldots} \approx 3</math>. | ||
+ | |||
+ | <math>f(6) - \sqrt[6]{\frac{7}{9} \cdot 10} = \sqrt[6]{\frac{70}{9}} = \sqrt[6]{7.\ldots} \approx 1</math>. | ||
+ | |||
+ | The final answer is therefore <math>2+1+1+3+1 = \boxed{\textbf{(A) }8}</math>. | ||
+ | |||
+ | ~KingRavi | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | In general, for positive integers <math>k</math> and real numbers <math>n > 10^{k}</math>, it is true that | ||
+ | <cmath>B(\sqrt[k]{\frac{n}{10^{k}}}) = B(\sqrt[k]{n}).</cmath> | ||
+ | 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 | ||
+ | <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> | ||
+ | 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>. | ||
+ | |||
+ | Similarly, we have | ||
+ | <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> | ||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~ihatemath123 | ||
+ | |||
+ | ==Solution 3 (Condensed Solution 1)== | ||
+ | Since <math>7777..7</math> is a <math>313</math> digit number and <math>\sqrt {7}</math> is around <math>2.5</math>, we have <math>f(2)</math> is <math>2</math>. <math>f(3)</math> is the same story, so <math>f(3)</math> is <math>1</math>. It is the same as <math>f(4)</math> as well, so <math>f(4)</math> is also <math>1</math>. However, <math>313</math> is <math>3</math> mod <math>5</math>, so we need to take the 5th root of <math>777</math>, which is between <math>3</math> and <math>4</math>, and therefore, <math>f(5)</math> is <math>3</math>. <math>f(6)</math> is the same as <math>f(4)</math>, since it is <math>1</math> more than a multiple of <math>6</math>. Therefore, we have <math>2+1+1+3+1</math> which is <math>\boxed {\textbf{(A)8}}</math>. | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | == Solution 4 == | ||
+ | First, we compute <math>f \left( 2 \right)</math>. | ||
+ | |||
+ | Because <math>N > 4 \cdot 10^{312}</math>, <math>\sqrt{N} > 2 \cdot 10^{156}</math>. | ||
+ | Because <math>N < 9 \cdot 10^{312}</math>, <math>\sqrt{N} < 3 \cdot 10^{156}</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) | ||
+ | |||
+ | ==Solution 5 (Guessing)== | ||
+ | |||
+ | Benford's Law states that in random numbers, the leading digit is more likely to be <math>1,2,</math> or <math>3</math> rather than <math>8</math> or <math>9</math>. From here, we can eliminate C, D, E. It is better to guess between A and B than not guess at all since your expected score from doing this is <math>3</math> points. | ||
+ | |||
+ | ~MathFun1000 | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/CqyVowe6qN4 | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==Video Solution 2 by WhyMath== | ||
+ | https://youtu.be/QEc1GhZnDuo | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2021 Fall|ab=B|num-a=20|num-b=18}} | ||
+ | {{MAA Notice}} |
Revision as of 13:16, 13 November 2023
Contents
Problem
Let be the positive integer , a -digit number where each digit is a . Let be the leading digit of the th root of . What is
Solution 1
We can rewrite as . 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 will be equal to the leading digit of .
Then is the first digit of
.
.
.
.
The final answer is therefore .
~KingRavi
Solution 2
For notation purposes, let be the number with digits, and let be the leading digit of . As an example, , because , and the first digit of that is .
Notice that for all numbers ; this is because , and dividing by does not affect the leading digit of a number. Similarly, In general, for positive integers and real numbers , it is true that Behind all this complex notation, all that we're really saying is that the first digit of something like has the same first digit as and .
The problem asks for
From our previous observation, we know that Therefore, . We can evaluate , the leading digit of , to be . Therefore, .
Similarly, we have Therefore, . We know , so .
Next, and , so .
We also have and , so .
Finally, and , so .
We have that .
~ihatemath123
Solution 3 (Condensed Solution 1)
Since is a digit number and is around , we have is . is the same story, so is . It is the same as as well, so is also . However, is mod , so we need to take the 5th root of , which is between and , and therefore, is . is the same as , since it is more than a multiple of . Therefore, we have which is .
~Arcticturn
Solution 4
First, we compute .
Because , . Because , .
Therefore, .
Second, we compute .
Because , . Because , .
Therefore, .
Third, we compute .
Because , . Because , .
Therefore, .
Fourth, we compute .
Because , . Because , .
Therefore, .
Fifth, we compute .
Because , . Because , .
Therefore, .
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 5 (Guessing)
Benford's Law states that in random numbers, the leading digit is more likely to be or rather than or . From here, we can eliminate C, D, E. It is better to guess between A and B than not guess at all since your expected score from doing this is points.
~MathFun1000
Video Solution by Interstigation
~Interstigation
Video Solution 2 by WhyMath
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.