2020 AMC 10B Problem 22

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

A divisor of the form $x + \sqrt{2x} + 1$ is a hint that it came from factoring a sum of squares by the Sophie Germain trick: $a^2 + b^2 = (a + b)^2 - 2ab$, and when $2ab$ is a perfect square you get a difference of squares. Here $a = 2^{101}, b = 1$ and $2ab = 2^{102} = (2^{51})^2$. Once the divisor is known to divide $2^{202} + 1$, the remainder question collapses to $-1 + 202$.

Test the pattern with small numbers: $2^{3} + 2^{2} + 1 = 13$ divides $2^{6} + 1 = 65 = 13 \cdot 5$, and $2^{5} + 2^{3} + 1 = 41$ divides $2^{10} + 1 = 1025 = 41 \cdot 25$. In general $(2^{2k+1} + 2^{k+1} + 1) \mid (2^{4k+2} + 1)$; the problem is $k = 50$. Small cases like these also let you confirm the remainder rule $2^{4k+2} \equiv -1$.

• Trying to reduce 2^202 by finding the order of 2 modulo the divisor, or forgetting to check that 201 is smaller than the divisor.
• Sign error in the expansion, concluding $2^{202} \equiv 1$ and answering $203$ (not a choice) or guessing $202$.