2020 AMC 10A Problem 22

First see how the three floors compare. Try an example: $n = 7$. Then $\frac{998}{7} \approx 142.6$, $\frac{999}{7} \approx 142.7$, $\frac{1000}{7} \approx 142.9$. All three floors are $142$, and the sum $3 \times 142$ is a multiple of $3$.

1. When does the floor go up? As the top goes $998 \to 999 \to 1000$, the floor of (top $\div n$) goes up by $1$ only when the top hits a multiple of $n$. So the three floors are all equal unless $n$ divides $999$ or $n$ divides $1000$.
2. Case: $n$ divides neither. All three floors equal some $q$. The sum is $3q$, a multiple of $3$. Not counted.
3. Case: $n$ divides $999$ but not $1000$. Example $n = 37$: floors $26, 27, 27$. The floor steps up at $999$, so the floors are $q - 1, q, q$ and the sum is $3q - 1$. Not a multiple of $3$. Counted.
4. Case: $n$ divides $1000$ but not $999$. Example $n = 8$: floors $124, 124, 125$. The sum is $3q - 2$. Not a multiple of $3$. Counted.
5. Case: $n$ divides both. Two numbers in a row share no common factor except $1$, so $n = 1$. The sum is $998 + 999 + 1000 = 2997 = 3 \times 999$, a multiple of $3$. Not counted.
6. Count the divisors. $999 = 3^3 \times 37$ has $4 \times 2 = 8$ divisors. $1000 = 2^3 \times 5^3$ has $4 \times 4 = 16$ divisors. Leave $n = 1$ out of each list: $7 + 15 = 22$. All of these are at most $1000$, as the problem requires.

The answer is $\boxed{\textbf{(A)}\ 22}$.

Floors of nearby fractions with the same bottom are almost always equal, and three equal numbers always add to a multiple of $3$. So the only interesting $n$ are the ones where the floor "steps up", and that happens exactly at divisors. Then it is a divisor count, with one special number ($n = 1$) to watch.

As $m$ increases by $1$, $\lfloor m/n \rfloor$ increases only when $m$ becomes a multiple of $n$. So the three floors for $m = 998, 999, 1000$ are all equal unless $n$ divides $999$ or $1000$.

Case 1: $n$ divides neither. All three floors equal some $q$, the sum is $3q$, divisible by $3$. Not counted.

Case 2: $n \mid 999$ but $n \nmid 1000$. Then $\lfloor 999/n \rfloor = \lfloor 1000/n \rfloor = q$ and $\lfloor 998/n \rfloor = q - 1$. Sum $3q - 1$: not divisible by $3$. Counted.

Case 3: $n \mid 1000$ but $n \nmid 999$. Then $\lfloor 1000/n \rfloor = q$ and the other two are $q - 1$. Sum $3q - 2$: counted.

Case 4: $n$ divides both $999$ and $1000$. Consecutive integers are coprime, so only $n = 1$; the sum $998 + 999 + 1000 = 2997 = 3 \cdot 999$ is divisible by $3$. Not counted.

Now count. $999 = 3^3 \cdot 37$ has $4 \cdot 2 = 8$ divisors; $1000 = 2^3 \cdot 5^3$ has $4 \cdot 4 = 16$. Removing $n = 1$ from each list leaves $7 + 15 = 22$ values, all at most $1000$.

The answer is $\boxed{\textbf{(A)}\ 22}$.