Difference between revisions of "2020 AMC 10B Problems/Problem 22"
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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>. | ||
− | ==Solution 6== | + | ==Solution 6 (Modular Arithmetic)== |
Let <math>n=2^{101}+2^{51}+1</math>. Then, <math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n} \equiv 2^{102}+2^{52}+203 \pmod{n} \equiv | Let <math>n=2^{101}+2^{51}+1</math>. Then, <math>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n} \equiv 2^{102}+2^{52}+203 \pmod{n} \equiv | ||
2(-2^{51}-1)+2^{52}+203 \pmod{n} \equiv 201 \pmod{n}</math>. Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>. | 2(-2^{51}-1)+2^{52}+203 \pmod{n} \equiv 201 \pmod{n}</math>. Thus, the remainder is <math>\boxed{\textbf{(D) } 201}</math>. | ||
− | ~ Leo.Euler | + | ~ Leo.Euler |
+ | |||
==Video Solutions== | ==Video Solutions== | ||
Revision as of 19:36, 19 August 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
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
.
Solution 6 (Modular Arithmetic)
Let . Then,
. Thus, the remainder is
.
~ Leo.Euler
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.