2022 AMC 10B Problem 17

The primes less than $10$ are $2, 3, 5, 7$. All five numbers are odd, so we look for factors of $3$, $5$ or $7$ using two facts: $a - b$ divides $a^n - b^n$ for every $n$, and $a + b$ divides $a^n + b^n$ when $n$ is odd.

• (A) $2^{606} - 1 = 4^{303} - 1$ is divisible by $4 - 1 = 3$.
• (B) $2^{606} + 1 = 4^{303} + 1$, and $303$ is odd, so it is divisible by $4 + 1 = 5$.
• (D) $2^{607} + 1$ with $607$ odd is divisible by $2 + 1 = 3$.
• (E) $2^{607} + 3^{607}$ with $607$ odd is divisible by $2 + 3 = 5$.

That leaves (C). To confirm, check $2^{607} - 1$ directly. Mod $3$: $2 \equiv -1$, so $2^{607} - 1 \equiv -1 - 1 \equiv 1$. Mod $5$: $2^4 \equiv 1$ and $607 = 4 \cdot 151 + 3$, so $2^{607} \equiv 2^3 = 8 \equiv 3$ and $2^{607} - 1 \equiv 2$. Mod $7$: $2^3 \equiv 1$ and $607 = 3 \cdot 202 + 1$, so $2^{607} - 1 \equiv 2 - 1 = 1$. None of $3, 5, 7$ divides it, and it is odd.

The answer is $\boxed{\textbf{(C)}\ 2^{607}-1}$.

Huge powers are never meant to be computed; the exponent's parity and factorization decide which algebraic identity applies. Even exponents let you rewrite $2^{606}$ as $4^{303}$ and use $4 \pm 1$; odd exponents give $a + b$ as a factor of $a^n + b^n$. When exactly one choice survives, the confirming residue check is short because $2$ has tiny cycles modulo $3$, $5$, $7$.

• Trying to check 2^607 − 1 directly for divisibility by 2, 3, 5, 7 before eliminating the others, and getting lost in exponent cycles.
• Applying "$a + b$ divides $a^n + b^n$" with an even $n$ (it fails), for instance believing $2^{606} + 1$ is divisible by $3$.