2020 AMC 10B Problems/Problem 22

Revision as of 18:09, 7 February 2020 by Advancedjus (talk | contribs) (Solution 2)

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

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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