Difference between revisions of "2022 AMC 10B Problems/Problem 25"
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<math>\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15</math> | <math>\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | |||
− | Then since <math>S_n</math> is the modular inverse of 7 in <math>\mathbb{Z}_{2^n}</math> , we can perform the Euclidean | + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
+ | |||
+ | ==Solution 2== | ||
+ | First, notice that <cmath>x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.</cmath> | ||
+ | Then since <math>S_n</math> is the modular inverse of <math>7</math> in <math>\mathbb{Z}_{2^n}</math>, we can perform the Euclidean Algorithm to find it for <math>n = 2019,2023</math>. | ||
− | Starting with <math>2019</math>, <cmath>7S_{2019} \equiv 1 \pmod{2^{2019}} | + | Starting with <math>2019</math>, |
− | + | <cmath>\begin{align*} | |
− | Now, take both sides <math>\ | + | 7S_{2019} &\equiv 1 \pmod{2^{2019}} \\ |
− | <cmath>0 \equiv 2^{2019}k + 1 \pmod{7}</cmath> | + | 7S_{2019} &= 2^{2019}k + 1. |
− | Using Fermat's Little Theorem, <cmath>2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}</cmath> | + | \end{align*}</cmath> |
− | Thus, <cmath>0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6</cmath> | + | Now, take both sides <math>\operatorname{mod} \ 7</math>: |
− | Therefore, <cmath>7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}</cmath> | + | <cmath>0 \equiv 2^{2019}k + 1 \pmod{7}.</cmath> |
+ | Using Fermat's Little Theorem, <cmath>2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}.</cmath> | ||
+ | Thus, <cmath>0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6.</cmath> | ||
+ | Therefore, <cmath>7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}.</cmath> | ||
− | We may repeat this same calculation with <math>S_{2023}</math> to yield <cmath>S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}</cmath> | + | We may repeat this same calculation with <math>S_{2023}</math> to yield <cmath>S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}.</cmath> |
Now, we notice that <math>S_n</math> is basically an integer expressed in binary form with <math>n</math> bits. | Now, we notice that <math>S_n</math> is basically an integer expressed in binary form with <math>n</math> bits. | ||
− | This gives rise to a simple inequality, <cmath>0 \leqslant S_n \leqslant 2^n</cmath> | + | This gives rise to a simple inequality, <cmath>0 \leqslant S_n \leqslant 2^n.</cmath> |
− | Since the maximum possible number that can be generated with <math>n</math> bits is <cmath>\underbrace{{11111\dotsc1}_2}_{n} = \sum_{k=0}^{n-1} 2^k = 2^n - 1 \leqslant 2^n</cmath> | + | Since the maximum possible number that can be generated with <math>n</math> bits is <cmath>\underbrace{{11111\dotsc1}_2}_{n} = \sum_{k=0}^{n-1} 2^k = 2^n - 1 \leqslant 2^n.</cmath> |
− | Looking at our calculations for <math>S_{2019}</math> and <math>S_{2023}</math>, we see that the only valid integers that satisfy that constraint are <math>j = h = 0</math> | + | Looking at our calculations for <math>S_{2019}</math> and <math>S_{2023}</math>, we see that the only valid integers that satisfy that constraint are <math>j = h = 0</math>. |
− | <cmath>\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{\textbf{(A)} | + | <cmath>\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{\textbf{(A) } 6}.</cmath> |
~ <math>\color{magenta} zoomanTV</math> | ~ <math>\color{magenta} zoomanTV</math> | ||
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==Solution 3 == | ==Solution 3 == | ||
− | As in Solution | + | As in Solution 2, we note that |
− | |||
<cmath>x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}.</cmath> | <cmath>x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}.</cmath> | ||
− | |||
We also know that <math>7S_{2023} \equiv 1 \pmod{2^{2023}}</math> and <math>7S_{2019} \equiv 1 \pmod{2^{2019}}</math>, this implies: | We also know that <math>7S_{2023} \equiv 1 \pmod{2^{2023}}</math> and <math>7S_{2019} \equiv 1 \pmod{2^{2019}}</math>, this implies: | ||
− | + | <cmath> \textbf{(1) } 7S_{2023}=2^{2023}\cdot{x} + 1, </cmath> | |
− | <cmath> \textbf{(1) } 7S_{2023}=2^{2023}\cdot{x} + 1 </cmath> | ||
<cmath> \textbf{(2) } 7S_{2019}=2^{2019}\cdot{y} + 1. </cmath> | <cmath> \textbf{(2) } 7S_{2019}=2^{2019}\cdot{y} + 1. </cmath> | ||
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Dividing by <math>7</math>, we can isolate the previous sums: | Dividing by <math>7</math>, we can isolate the previous sums: | ||
− | + | <cmath> \textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7}, </cmath> | |
− | <cmath> \textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7} </cmath> | ||
<cmath> \textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7}. </cmath> | <cmath> \textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7}. </cmath> | ||
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The maximum value of <math>S_n</math> occurs when every <math>x_i</math> is equal to <math>1</math>. Even when this happens, the value of <math>S_n</math> is less than <math>2^n</math>. Therefore, we can construct the following inequalities: | The maximum value of <math>S_n</math> occurs when every <math>x_i</math> is equal to <math>1</math>. Even when this happens, the value of <math>S_n</math> is less than <math>2^n</math>. Therefore, we can construct the following inequalities: | ||
− | + | <cmath> \textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7} < 2^{2023}, </cmath> | |
− | <cmath> \textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7} < 2^{2023} </cmath> | ||
<cmath> \textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7} < 2^{2019}. </cmath> | <cmath> \textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7} < 2^{2019}. </cmath> | ||
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From these two equations, we can deduce that both <math>x</math> and <math>y</math> are less than <math>7</math>. | From these two equations, we can deduce that both <math>x</math> and <math>y</math> are less than <math>7</math>. | ||
Reducing <math>\textbf{1}</math> and <math>\textbf{2}</math> <math>\pmod{7},</math> we see that | Reducing <math>\textbf{1}</math> and <math>\textbf{2}</math> <math>\pmod{7},</math> we see that | ||
− | + | <cmath>2^{2023}\cdot{x}\equiv 6\pmod{7},</cmath> | |
− | <cmath>2^{2023}\cdot{x}\equiv 6\pmod{7}</cmath> | ||
and | and | ||
<cmath>2^{2019}\cdot{y}\equiv 6\pmod{7}.</cmath> | <cmath>2^{2019}\cdot{y}\equiv 6\pmod{7}.</cmath> | ||
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Therefore, <math>2^{2023}\equiv 2 \pmod 7</math> and <math>2^{2019} \equiv 1 \pmod {7}.</math> Substituing this back into the above equations, | Therefore, <math>2^{2023}\equiv 2 \pmod 7</math> and <math>2^{2019} \equiv 1 \pmod {7}.</math> Substituing this back into the above equations, | ||
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<cmath>2x\equiv{6}\pmod{7}</cmath> | <cmath>2x\equiv{6}\pmod{7}</cmath> | ||
and | and | ||
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The requested sum is | The requested sum is | ||
− | + | <cmath>\begin{align*} | |
− | <cmath>\frac{S_{2023}-S_{2019}}{2^{2019}} = \frac{\frac{2^{2023}\cdot{x} + 1}{7} - \frac{2^{2019}\cdot{y} + 1}{7}}{2^{2019}} | + | \frac{S_{2023}-S_{2019}}{2^{2019}} &= \frac{\frac{2^{2023}\cdot{x} + 1}{7} - \frac{2^{2019}\cdot{y} + 1}{7}}{2^{2019}} \\ |
− | + | &= \frac{1}{2^{2019}}\left(\frac{2^{2023}\cdot{3} + 1}{7} -\left(\frac{2^{2019}\cdot{6} + 1}{7} \right)\right) \\ | |
− | + | &= \frac{3\cdot{2^4}-6}{7} \\ | |
− | + | &= \boxed{\textbf{(A) } 6}. | |
− | + | \end{align*}</cmath> | |
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-Benedict T (countmath1) | -Benedict T (countmath1) | ||
Revision as of 08:48, 5 December 2022
- The following problem is from both the 2022 AMC 10B #25 and 2022 AMC 12B #23, so both problems redirect to this page.
Contents
Problem
Let be a sequence of numbers, where each
is either
or
. For each positive integer
, define
Suppose for all
. What is the value of the sum
Solution 1
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
First, notice that
Then since
is the modular inverse of
in
, we can perform the Euclidean Algorithm to find it for
.
Starting with ,
Now, take both sides
:
Using Fermat's Little Theorem,
Thus,
Therefore,
We may repeat this same calculation with to yield
Now, we notice that
is basically an integer expressed in binary form with
bits.
This gives rise to a simple inequality,
Since the maximum possible number that can be generated with
bits is
Looking at our calculations for
and
, we see that the only valid integers that satisfy that constraint are
.
~
Solution 3
As in Solution 2, we note that
We also know that
and
, this implies:
Dividing by
, we can isolate the previous sums:
The maximum value of
occurs when every
is equal to
. Even when this happens, the value of
is less than
. Therefore, we can construct the following inequalities:
From these two equations, we can deduce that both
and
are less than
.
Reducing and
we see that
and
The powers of repeat every
Therefore, and
Substituing this back into the above equations,
and
Since and
are integers less than
, the only values of
and
are
and
respectively.
The requested sum is
-Benedict T (countmath1)
Video Solutions
~ ThePuzzlr
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by OmegaLearn Using Binary and Modular Arithmetic
~ pi_is_3.14
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last problem | |
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 |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.