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

Revision as of 11:03, 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

If $x$ = $2^(50)$ , $4x^{4} +$ 202 $=$ ( $2x^{2} +$ 2x $-$ 1 $)($ 2x^{2} + $2x +$ 1 $) +$ 201 $, So the remainder is$ 201-(D)$$ (Error compiling LaTeX. Unknown error_msg)

@@ Line 39: / Line 39: @@
 ==Solution 3==
-If <math>x</math> = <cmath>2^(50)</cmath>, <cmath>4x^{4} + 202 = </cmath>(<cmath>2x^{2} + </cmath>2x - 1)(<cmath>2x^{2} + </cmath>2x + <cmath>1</cmath>) + <cmath>201</cmath>,
+If <math>x</math> = <cmath>2^(50)</cmath>, <cmath>4x^{4} + </cmath>202<cmath> = </cmath>(<cmath>2x^{2} + </cmath>2x<cmath> - </cmath>1<cmath>)(</cmath>2x^{2} + <cmath>2x + </cmath>1<cmath>) + </cmath>201<cmath>,
-So the remainder is 201-(D)
+So the remainder is </cmath>201-(D)<math></math>
 ==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
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