2020 AMC 10B Problems/Problem 22
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 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 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)
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.