Difference between revisions of "2020 AMC 10B Problems/Problem 22"
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==Solution== | ==Solution== | ||
+ | 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 denominator 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 | ||
+ | |||
+ | <cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath> | ||
+ | |||
+ | Rearranging, we can see that this is exactly what we need: | ||
+ | |||
+ | <cmath>\frac{4x^4+1}{2x^2+2x+1}=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 | ||
+ | |||
+ | Note: You could take inputs on a computer and get the remainder by doing (2^202 + 202) % (2^201 + 2^51 + 1). This ends up being 201(Informatics Olympiad) | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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== | ||
+ | |||
+ | <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> | ||
+ | |||
+ | Thus, we see that the remainder is surely <math>\boxed{\textbf{(D) } 201}</math> | ||
+ | |||
+ | (Source: https://artofproblemsolving.com/community/c5h2001950p14000817) | ||
+ | So | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | 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}+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== | ||
+ | https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:03, 28 June 2020
Contents
Problem
What is the remainder when is divided by ?
Solution
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
Note: You could take inputs on a computer and get the remainder by doing (2^202 + 202) % (2^201 + 2^51 + 1). This ends up being 201(Informatics Olympiad)
Solution 2
Similar to Solution 1, let . It suffices to find remainder of . Dividing polynomials results in a remainder of .
MAA Original Solution
Thus, we see that the remainder is surely
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817) So
Solution 3
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 .
Video Solution
https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx
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 |
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