Difference between revisions of "2020 AMC 10B Problems/Problem 22"
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==Solution 1 (MAA Original Solution)== | ==Solution 1 (MAA Original Solution)== | ||
+ | Completing the square, then difference of squares: | ||
<cmath> | <cmath> | ||
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\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | |||
− | Thus, we see that the remainder is | + | |
+ | Thus, we see that the remainder is <math>\boxed{\textbf{(D) } 201}</math> | ||
(Source: https://artofproblemsolving.com/community/c5h2001950p14000817) | (Source: https://artofproblemsolving.com/community/c5h2001950p14000817) | ||
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==Solution 4 == | ==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>. | 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>. | ||
+ | |||
+ | 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)== | ==Solution 5 (Modular Arithmetic)== | ||
− | Let <math>n=2^{101}+2^{51}+1</math>. Then, | + | 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>2^{202}+202 \equiv (-2^{51}-1)^2 + 202 </math> |
− | <math>\equiv 2^{102}+2^{52}+203 | + | <math>\equiv 2^{102}+2^{52}+203 </math> |
<math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. | <math>= 2(n-1)+203 \equiv 201 \pmod{n}</math>. |
Revision as of 16:04, 8 June 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.