# Difference between revisions of "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

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

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

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

## See Also

 2020 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 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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