Difference between revisions of "2020 AMC 10B Problems/Problem 22"
(→MAA Original Solution) |
|||
(33 intermediate revisions by 15 users not shown) | |||
Line 5: | Line 5: | ||
<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>. | ||
Line 23: | Line 23: | ||
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 2 (MAA Original Solution)== | |
− | + | <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 surely <math>\boxed{\textbf{(D) } 201}</math> |
+ | |||
+ | (Source: https://artofproblemsolving.com/community/c5h2001950p14000817) | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ==Solution 5 (Modular Arithmetic)== | ||
+ | Let <math>n=2^{101}+2^{51}+1</math>. Then, | ||
+ | |||
+ | <math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}</math> | ||
− | < | + | <math>\equiv 2^{102}+2^{52}+203 \pmod{n}</math> |
− | Thus, | + | <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) | ||
+ | |||
+ | ==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=== | ||
+ | https://youtu.be/gPqd-yKQdFg | ||
− | + | ===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== | ==See Also== |
Revision as of 20:43, 31 October 2021
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 (MAA Original Solution)
Thus, we see that the remainder is surely
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817)
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 .
Solution 5 (Modular Arithmetic)
Let . Then,
.
Thus, the remainder is .
~ Leo.Euler
~ (edited by asops)
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.