2020 AMC 10A Problem 22

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 $\boxed{\textbf{(A)}\ 22}$.

Three consecutive numerators over the same denominator have floors that differ by at most $1$, and the sum of three equal integers is automatically a multiple of $3$. The whole problem is therefore about where the floor "steps," which happens exactly at divisors. Then the count is a divisor-counting exercise with one edge case.

• Adding the divisor counts 8 + 16 = 24 without handling n = 1, which divides both 999 and 1000 and actually makes the sum divisible by 3.
• Subtracting $1$ only once for the double-counted $n = 1$ (giving $23$) instead of removing it entirely, or forgetting that $998$ has no special role since $\lfloor 998/n \rfloor$ is never the largest floor.