Difference between revisions of "2020 AMC 10B Problems/Problem 22"
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https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx | https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx | ||
− | ==Video Solution 2== | + | ==Video Solution 2 by the BeautyOfMath== |
https://youtu.be/gPqd-yKQdFg | https://youtu.be/gPqd-yKQdFg | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}} | {{AMC10 box|year=2020|ab=B|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:00, 4 December 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
Video Solution 2 by the BeautyOfMath
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.