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

(Created page with "==Problem== What is the remainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>? <math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(...")
 
(→‎Solution)
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==Solution==
 
==Solution==
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Let <math>x=2^{50}</math>. We are now looking for the remainder of <math>\frac{4x^4+202}{2x^2+2x+1}</math>.
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We could proceed with polynomial division, but the denominator looks awfully similar to the [[Sophie Germain Identity]],  which states that <cmath>a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)</cmath>
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Let's use the identity, with <math>a=1</math> and <math>b=x</math>, so we have
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<cmath>1+4x^4=(1+2x^2+2x)(1+2x^2-2x)</cmath>.
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Rearranging, we can see that this is exactly what we need:
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<cmath>\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1</cmath>.
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So <cmath>\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1} </cmath>.
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Since the first half divides cleanly as shown earlier, the remainder must be <math>\boxed\textbf{(D) }201}</math> ~quacker88
  
 
==See Also==  
 
==See Also==  

Revision as of 16:47, 7 February 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 $\boxed\textbf{(D) }201}$ (Error compiling LaTeX. Unknown error_msg) ~quacker88

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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