# 2020 AMC 10B Problems/Problem 22

## 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 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 numerator 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 (MAA Original Solution) \begin{align*} 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. \end{align*}

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

## Solution 3

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 4

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

## Solution 5 (Modular Arithmetic)

Let $n=2^{101}+2^{51}+1$. Then, $2^{202}+202 \equiv (-2^{51}-1)^2 + 202 \pmod{n}$ $\equiv 2^{102}+2^{52}+203 \pmod{n}$ $= 2(n-1)+203 \equiv 201 \pmod{n}$.

Thus, the remainder is $\boxed{\textbf{(D) } 201}$.

~ Leo.Euler

~ (edited by asops)

## Solution 6(Author: Shiva Kannan - Least insightful & very straightforward + Manipulation)

We can repeatedly manipulate the numerator to make parts of it divisible by the denominator: $$\frac{2^{202}+202}{2^{101}+2^{51}+1}$$ $$= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}$$ $$= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}$$ $$=2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}$$ $$=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1}$$ $$= 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1}$$ $$= 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.$$

Clearly, $201 < 2^{201} + 2^{51} + 1$, hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is $\boxed{\textbf{(D) } 201}$.

## Video Solutions

### Video Solution 4 Using Sophie Germain's Identity

~ pi_is_3.14

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