Sum of Floors (2020 AMC 10A #22)

2020 AMC 10A #22

Problem: For how many positive integers $n \leq 1000$ is $$\lfloor{\frac{998}{n}}\rfloor + \lfloor{\frac{999}{n}}\rfloor + \lfloor{\frac{1000}{n}}\rfloor$$

not divisible by 3? (Recall that $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.)

Solution: Observe that the 3 values $\lfloor{\frac{998}{n}}\rfloor$, $\lfloor \frac{999}{n} \rfloor$, and $\lfloor{\frac{1000}{n}}\rfloor$ must have exactly 1 of them that is not equal to the others, to satisfy the condition that the expression 

$$\lfloor{\frac{998}{n}}\rfloor + \lfloor{\frac{999}{n}}\rfloor + \lfloor{\frac{1000}{n}}\rfloor$$

is not divisible by 3.

Note that the "turning points", or values of n that produce different values for $\lfloor{\frac{998}{n}}\rfloor$, $\lfloor \frac{999}{n} \rfloor$, and $\lfloor{\frac{1000}{n}}\rfloor$ are the factors of 999 and 1000.

We realize that 998 has 4 factors: 1, 2, 499, 998. Note that when $n=1$, 499, and 998, they all have the sum that is divisible by 3. However, when $n=2$, which is also a factor of 1000, we have 1498, which is not divisible by 3. 

Next, we find that 999 has 7 factors (not including 1) and 1000 has 15 factors (not including 1) for a total of 22 values of $n$ that meets the condition.