Difference between revisions of "2020 AMC 10B Problems/Problem 22"
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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 | ||
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==Solution 2== | ==Solution 2== |
Revision as of 18:40, 28 November 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
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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