2018 AMC 10B Problem 21

Break $323$ into primes first: $323 = 17 \times 19$.

1. Why the choices $324$ and $330$ fail. Suppose the next divisor $d$ has no factor $17$ or $19$. Then $n$ is a multiple of $323$ and also of $d$, with no factors in common, so $n$ is a multiple of $323 \times d$. But
$$323 \times 324 = 104652,$$
is already six digits, and $d$ is at least $324$. So $n$ could not have four digits.
2. So $d$ is a multiple of $17$ or $19$. Neither $324 = 2^2 \times 3^4$ nor $330 = 2 \times 3 \times 5 \times 11$ is, so both are ruled out.
3. Smallest such $d$ above 323. Since $323 = 17 \times 19$:
- the next multiple of $17$ is $17 \times 20 = 340$;
- the next multiple of $19$ is $19 \times 18 = 342$.

So $d$ is at least $340$.
4. Show $340$ really happens. Try $n = 17 \times 19 \times 20 = 6460$. It is even and has four digits, and both $323$ and $340$ divide it. Its prime factors are $2 \times 2 \times 5 \times 17 \times 19$. Could a divisor land between $323$ and $340$? A divisor using both $17$ and $19$ is $323$ times $1, 2, 4, 5, \ldots$, so the next one is $646$. One using $17$ but not $19$ is $17$ times a divisor of $20$, the closest being $17 \times 20 = 340$. One using $19$ but not $17$ is at most $19 \times 20 = 380$ and otherwise $19 \times 10 = 190$ or less. One using neither is at most $20$. So nothing lies strictly between, and $340$ comes right after $323$.

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

Two divisors of the same number that share no prime factor force the number to be a multiple of their product. With a four-digit limit and divisors around $300$, that product is far too big, so the next divisor must share a prime with $323$. That leaves only a couple of candidates, and an explicit example confirms the smallest one works.