Difference between revisions of "2020 AMC 10B Problems/Problem 22"
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<math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math> | <math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math> | ||
− | ==Solution 1== | + | ==Solution 1 (MAA Original Solution)== |
+ | Completing the square, then difference of squares: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\ | ||
+ | &= (2^{101} + 1)^2 - 2^{102} + 201\\ | ||
+ | &= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | |||
+ | Thus, we see that the remainder is <math>\boxed{\textbf{(D) } 201}</math> | ||
+ | |||
+ | (Source: https://artofproblemsolving.com/community/c5h2001950p14000817) | ||
+ | |||
+ | ==Solution 2== | ||
Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>. | Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>. | ||
− | We could proceed with polynomial division, but the | + | We could proceed with polynomial division, but the numerator looks awfully similar to the [[Sophie Germain Identity]], which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath> |
Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have | Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have | ||
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So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath> | So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath> | ||
− | Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> ~quacker88 | + | Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed{\textbf{(D) }201}</math> |
+ | |||
+ | ~quacker88 | ||
+ | |||
+ | |||
+ | |||
+ | ==Solution 3 (Same As Solution 2)== | ||
+ | We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>. | ||
+ | Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>. | ||
+ | We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math> | ||
+ | |||
+ | ==Solution 4 == | ||
+ | We let <math>x=2^{50.5}</math>. That means <math>2^{202}+202=x^{4}+202</math> and <math>2^{101}+2^{51}+1=x^{2}+2^{0.5}x+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>. | ||
+ | |||
+ | The long division is essentially the same if you work with <math>x=2</math>, or do repeated multiplication and subtraction using the original expression. | ||
+ | |||
+ | ==Solution 5 (Modular Arithmetic)== | ||
+ | Let <math>n=2^{101}+2^{51}+1</math>. Then, mod <math>n</math>: | ||
+ | |||
+ | <math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 </math> | ||
+ | |||
+ | <math>\equiv 2^{102}+2^{52}+203 </math> | ||
+ | |||
+ | <math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. | ||
+ | |||
+ | Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>. | ||
+ | |||
+ | ~ Leo.Euler | ||
− | + | ~ (edited by asops) | |
− | |||
− | ==Solution | + | ==Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)== |
− | + | We can repeatedly manipulate the numerator to make parts of it divisible by the denominator: | |
− | + | <cmath> \frac{2^{202}+202}{2^{101}+2^{51}+1} </cmath> <cmath>= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</cmath> <cmath>= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}</cmath> <cmath>=2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}</cmath> <cmath>=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1}</cmath> <cmath>= 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1}</cmath> <cmath>= 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.</cmath> | |
− | ( | + | Clearly, <math>201 < 2^{201} + 2^{51} + 1</math>, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is <math>\boxed{\textbf{(D) } 201}</math>. |
+ | ==Video Solutions== | ||
− | ==Solution | + | ===Video Solution 1 by Mathematical Dexterity (2 min)=== |
− | + | https://www.youtube.com/watch?v=lLWURnmpPQA | |
− | |||
− | |||
− | ==Solution | + | ===Video Solution 2 by The Beauty Of Math=== |
− | + | https://youtu.be/gPqd-yKQdFg | |
− | ==Video Solution | + | ===Video Solution 3=== |
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx | https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx | ||
− | ==Video Solution | + | ===Video Solution 4 Using Sophie Germain's Identity=== |
− | https://youtu.be/ | + | https://youtu.be/ba6w1OhXqOQ?t=5155 |
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 21:51, 16 August 2024
Contents
Problem
What is the remainder when is divided by
?
Solution 1 (MAA Original Solution)
Completing the square, then difference of squares:
Thus, we see that the remainder is
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)
Solution 2
Let . We are now looking for the remainder of
.
We could proceed with polynomial division, but the numerator looks awfully similar to the Sophie Germain Identity, which states that
Let's use the identity, with and
, so we have
Rearranging, we can see that this is exactly what we need:
So
Since the first half divides cleanly as shown earlier, the remainder must be
~quacker88
Solution 3 (Same As Solution 2)
We let and
.
Next we write
.
We know that
by the Sophie Germain identity so to find
we find that
which shows that the remainder is
Solution 4
We let . That means
and
. Then, we simply do polynomial division, and find that the remainder is
.
The long division is essentially the same if you work with , or do repeated multiplication and subtraction using the original expression.
Solution 5 (Modular Arithmetic)
Let . Then, mod
:
.
Thus, the remainder is .
~ Leo.Euler
~ (edited by asops)
Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)
We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:
Clearly, , hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is
.
Video Solutions
Video Solution 1 by Mathematical Dexterity (2 min)
https://www.youtube.com/watch?v=lLWURnmpPQA
Video Solution 2 by The Beauty Of Math
Video Solution 3
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx
Video Solution 4 Using Sophie Germain's Identity
https://youtu.be/ba6w1OhXqOQ?t=5155
~ pi_is_3.14
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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.