Difference between revisions of "2021 Fall AMC 10B Problems/Problem 19"
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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> | 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 == | + | ==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>. | ||
+ | |||
+ | 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>. | 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>. | ||
Line 37: | Line 57: | ||
~ihatemath123 | ~ihatemath123 | ||
− | ==Solution | + | ==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 <math> | + | 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 | ~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== | ==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}} |
Latest revision as of 14:16, 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.