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

Revision as of 20:08, 15 January 2021

Problem

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

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

Video Solution Using Sophie Germain's Identity

https://youtu.be/ba6w1OhXqOQ?t=5155

~ pi_is_3.14

Solution 1

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

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 $\boxed{\textbf{(D) } 201}$ .

Solution 3 (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)

Solution 4

We let $x = 2^{50}$ and $2^{202} + 202 = 4x^{4} + 202$ . Next we write $2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1$ . We know that $4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)$ by the Sophie Germain identity so to find $4x^{4} + 202,$ we find that $4x^{4} + 202 = 4x^{4} + 201 + 1$ which shows that the remainder is $\boxed{\textbf{(D) } 201}$

Solution 5

We let $x=2^{50.5}$ . That means $2^{202}+202=x^{4}+202$ and $2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1$ . Then, we simply do polynomial division, and find that the remainder is $\boxed{\textbf{(D) } 201}$ .

Video Solution 1 by The Beauty Of Math

https://youtu.be/gPqd-yKQdFg

Video Solution 2

https://www.youtube.com/watch?v=Qs6UnryIAI8&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=9&t=0s ~ MathEx

@@ Line 4: / Line 4: @@
 <math>\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202</math>
+== Video Solution Using Sophie Germain's Identity==
+https://youtu.be/ba6w1OhXqOQ?t=5155
+~ pi_is_3.14
 ==Solution 1==

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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