2018 AMC 10B Problem 21

Factor $323 = 17 \cdot 19$. Let $d > 323$ be the next divisor of $n$.

Both $323$ and $d$ divide $n$, so $\operatorname{lcm}(323, d)$ divides $n$ and hence is at most $9999$. If $d$ shared no prime factor with $323$, the lcm would be $323 d \ge 323 \cdot 324 > 100000$, impossible. So $d$ is a multiple of $17$ or of $19$.

The multiples of $17$ just above $323 = 17 \cdot 19$ start at $17 \cdot 20 = 340$; the multiples of $19$ start at $19 \cdot 18 = 342$. So $d \ge 340$, and this rules out $324$ and $330$ (neither is divisible by $17$ or $19$).

Now check that $d = 340$ is attainable: take
$$n = \operatorname{lcm}(323, 340) = 17 \cdot 19 \cdot 20 = 6460,$$
which is even and has four digits. Its divisors are products of $2^a 5^b 17^c 19^e$, and none of them lies strictly between $323$ and $340$, so $340$ really is the divisor written immediately after $323$.

The answer is $\boxed{\textbf{(C)}\ 340}$.

Two divisors of the same number cannot be "too coprime" relative to its size: their lcm must still fit inside $n$. A four-digit ceiling forces consecutive large divisors to share a prime factor, which converts a search over all $d$ into a search over multiples of $17$ or $19$. Always finish with an explicit $n$ to confirm the minimum is achieved.

Test the choices in order. $324 = 2^2 3^4$ and $330 = 2 \cdot 3 \cdot 5 \cdot 11$ are coprime to $323$, so $n$ would be a multiple of $323 \cdot 324$ or $323 \cdot 330$, both far beyond four digits. $340 = 17 \cdot 20$ works with $n = 6460$.

• Answering 324 (the next integer) or 330 without checking that n would have to be a multiple of lcm(323, d), which has five digits when d is coprime to 323.
• Jumping to $646 = 2 \cdot 323$ on the theory that the next divisor after $323$ must be a multiple of it.