2020 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 22 Hard

What is the remainder when 2^{202}+202 is divided by 2^{101}+2^{51}+1 ?(2020 AMC 10B, Question #22)

  • A.

    100

  • B.

    101

  • C.

    200

  • D.

    201

  • E.

    202

Answer:D

Solution 1： Let x=2^{50}. We are now looking for the remainder of \frac{4 x^{4}+202}{2 x^{2}+2 x+1} We could proceed with polynomial division, but the denominator looks awfully similar to the Sophie Germain Identity, which states that a^{4}+4 b^{4}=\left(a^{2}+2 b^{2}+2 a b\right)\left(a^{2}+2 b^{2}-2 a b\right) Let's use the identity, with a=1 and b=x, so we have 1+4 x^{4}=\left(1+2 x^{2}+2 x\right)\left(1+2 x^{2}-2 x\right) Rearranging, we can see that this is exactly what we need: \begin{aligned} &\frac{4 x^{4}+1}{2 x^{2}+2 x+1}=2 x^{2}-2 x+1 \\ &\text { So } \frac{4 x^{4}+202}{2 x^{2}+2 x+1}=\frac{4 x^{4}+1}{2 x^{2}+2 x+1}+\frac{201}{2 x^{2}+2 x+1} \end{aligned} Since the first half divides cleanly as shown earlier, the remainder must be (D) 201.

Solution 2:

Similar to Solution 1 , let x=2^{50}. It suffices to find remainder of \frac{4 x^{4}+202}{2 x^{2}+2 x+1}. of 2 x^{2}+2 x+1. Dividing polynomials results in a remainder of (D) 201

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