2020 AMC 10B Problem 22

1. Name the pieces. Let $a = 2^{101}$. Then $2^{202} = a^2$ and the divisor is $m = a + 2^{51} + 1$.
2. Why look for a product? If $2^{202}$ plus a small number turns out to be a multiple of $m$, the remainder is easy. The middle term is the clue: $(2^{51})^2 = 2^{102} = 2 \times 2^{101} = 2a$.
3. Multiply $m$ by its "partner" with the middle sign flipped. Group as $(a + 1) + 2^{51}$ and $(a + 1) - 2^{51}$ and use (sum)(difference) = difference of squares:
$$\bigl((a + 1) + 2^{51}\bigr)\bigl((a + 1) - 2^{51}\bigr) = (a + 1)^2 - (2^{51})^2 = a^2 + 2a + 1 - 2a = a^2 + 1.$$
4. So $2^{202} + 1 = a^2 + 1$ is a multiple of $m$.
5. Split the number we want.
$$2^{202} + 202 = (2^{202} + 1) + 201.$$
The first part is a multiple of $m$, so the remainder comes from $201$.
6. Check the size. $m$ is enormous (bigger than $2^{101}$), so $201$ is already smaller than $m$. The remainder is exactly $201$.

The answer is $\boxed{\textbf{(D)}\ 201}$.

A divisor that looks like $a + (\text{something}) + 1$, where the something squared equals $2a$, comes from factoring $a^2 + 1$ with a difference of squares. Once you know the divisor goes evenly into $2^{202} + 1$, the question becomes "what is left after taking away that multiple?", which is just $202 - 1 = 201$. In the language of remainders: $2^{202} \equiv -1 \pmod m$, meaning $2^{202}$ is one less than a multiple of $m$.

Test the pattern on small numbers. With $a = 2^3 = 8$: the divisor $8 + 2^2 + 1 = 13$, and $8^2 + 1 = 65 = 5 \times 13$. With $a = 2^5 = 32$: the divisor $32 + 2^3 + 1 = 41$, and $32^2 + 1 = 1025 = 25 \times 41$. The pattern holds, and our problem is the same with $a = 2^{101}$.

Let $m = 2^{101} + 2^{51} + 1$. The middle term $2^{51}$ satisfies $(2^{51})^2 = 2^{102} = 2 \cdot 2^{101}$, which suggests the difference-of-squares pairing
$$(2^{101} + 1 + 2^{51})(2^{101} + 1 - 2^{51}) = (2^{101} + 1)^2 - (2^{51})^2 = 2^{202} + 2\cdot 2^{101} + 1 - 2^{102} = 2^{202} + 1.$$

So $m$ divides $2^{202} + 1$, which means $2^{202} \equiv -1 \pmod m$. Then
$$2^{202} + 202 \equiv -1 + 202 = 201 \pmod m.$$

Since $0 \le 201 < m$, the remainder is exactly $201$.

The answer is $\boxed{\textbf{(D)}\ 201}$.