Difference between revisions of "2020 AMC 10B Problems/Problem 22"
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− | We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202 | + | We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202</cmath>. |
− | Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1 | + | Next we write <cmath>2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1</cmath>. |
We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math> | We know that <cmath>4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)</cmath> by the Sophie Germain identity so to find <cmath>4x^{4} + 202,</cmath> we find that <cmath>4x^{4} + 202 = 4x^{4} + 201 + 1</cmath> which shows that the remainder is <math>\boxed{\textbf{(D) } 201}</math> | ||
Revision as of 15:26, 6 April 2020
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
Note: You could take inputs on a computer and get the remainder by doing (2^202 + 202) % (2^201 + 2^51 + 1). This ends up being 201(Informatics Olympiad)
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
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.