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Difference between revisions of "2020 AMC 10B Problems/Problem 22"

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So
 
So
  
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==Solution 3==
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We let <cmath>x = 2^{50}</cmath> and <cmath>2^{202} + 202 = 4x^{4} + 202.</cmath>
  
 
==See Also==  
 
==See Also==  

Revision as of 12:17, 6 April 2020

Problem

What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$?

$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$

Solution

Let $x=2^{50}$. We are now looking for the remainder of $\frac{4x^4+202}{2x^2+2x+1}$.

We could proceed with polynomial division, but the denominator looks awfully similar to the Sophie Germain Identity, which states that \[a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)\]

Let's use the identity, with $a=1$ and $b=x$, so we have

\[1+4x^4=(1+2x^2+2x)(1+2x^2-2x)\]

Rearranging, we can see that this is exactly what we need:

\[\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1\]

So \[\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1}\]

Since the first half divides cleanly as shown earlier, the remainder must be $\boxed{\textbf{(D) }201}$ ~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 $x=2^{50}$. It suffices to find remainder of $\frac{4x^4+202}{2x^2+2x+1}$. Dividing polynomials results in a remainder of $\boxed{\textbf{(D) } 201}$.

MAA Original Solution

\[2^{202} + 202 = (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\]\[= (2^{101} + 1)^2 - 2^{102} + 201\] \[= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.\]

Thus, we see that the remainder is surely $\boxed{\textbf{(D) } 201}$

(Source: https://artofproblemsolving.com/community/c5h2001950p14000817) So


Solution 3

We let \[x = 2^{50}\] and \[2^{202} + 202 = 4x^{4} + 202.\]

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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