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== | + | ==Solution 1== |
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>. | ||
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Similar to Solution 1, let <math>x=2^{50}</math>. It suffices to find remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>. Dividing polynomials results in a remainder of <math>\boxed{\textbf{(D) } 201}</math>. | Similar to Solution 1, let <math>x=2^{50}</math>. It suffices to find remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>. Dividing polynomials results in a remainder of <math>\boxed{\textbf{(D) } 201}</math>. | ||
− | ==MAA Original Solution== | + | ==Solution 3 (MAA Original Solution)== |
<cmath>2^{202} + 202 = (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201</cmath><cmath>= (2^{101} + 1)^2 - 2^{102} + 201</cmath> <cmath>= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.</cmath> | <cmath>2^{202} + 202 = (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201</cmath><cmath>= (2^{101} + 1)^2 - 2^{102} + 201</cmath> <cmath>= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.</cmath> | ||
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(Source: https://artofproblemsolving.com/community/c5h2001950p14000817) | (Source: https://artofproblemsolving.com/community/c5h2001950p14000817) | ||
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− | ==Solution | + | ==Solution 4== |
We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>. | 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>. | 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> | 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 | + | ==Solution 5== |
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}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>. | 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}+x\sqrt{2}+1</math>. Then, we simply do polynomial division, and find that the remainder is <math>\boxed{\textbf{(D) } 201}</math>. | ||
− | ==Video Solution== | + | ==Video Solution 1== |
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 2 by the | + | ==Video Solution 2 by the Beauty Of Math== |
https://youtu.be/gPqd-yKQdFg | https://youtu.be/gPqd-yKQdFg | ||
Revision as of 11:10, 24 December 2020
Contents
Problem
What is the remainder when is divided by ?
Solution 1
Let . We are now looking for the remainder of .
We could proceed with polynomial division, but the denominator 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 2
Similar to Solution 1, let . It suffices to find remainder of . Dividing polynomials results in a remainder of .
Solution 3 (MAA Original Solution)
Thus, we see that the remainder is surely
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)
Solution 4
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 5
We let . That means and . Then, we simply do polynomial division, and find that the remainder is .
Video Solution 1
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx
Video Solution 2 by the Beauty Of Math
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.