AMC10 2020 A

Answer (A): If $n$ is a divisor of $999$, then $\left\lfloor \frac{999}{n}\right\rfloor = 1 + \left\lfloor \frac{998}{n}\right\rfloor$. Otherwise, $\left\lfloor \frac{999}{n}\right\rfloor = \left\lfloor \frac{998}{n}\right\rfloor$. Similarly, $\left\lfloor \frac{1000}{n}\right\rfloor = 1 + \left\lfloor \frac{999}{n}\right\rfloor$ if $n$ is a divisor of $1000$, and $\left\lfloor \frac{1000}{n}\right\rfloor = \left\lfloor \frac{999}{n}\right\rfloor$ otherwise. Therefore if $n$ divides neither $999$ nor $1000$, then the given sum has the form $a+a+a=3a$, so it is a multiple of $3$. Further, if $n$ divides both $999$ and $1000$ (that is, if $n=1$), then the sum has the form $a+(a+1)+(a+2)=3a+3$, which is also a multiple of $3$. Thus in order for the sum not to be a multiple of $3$, $n$ must be a divisor of either $999$ or $1000$, but not both, so that the sum has the form $a+a+(a+1)=3a+1$ or $a+(a+1)+(a+1)=3a+2$. Because $999=3^3\cdot 37^1$ and $1000=2^3\cdot 5^3$, it follows that $999$ has $(3+1)(1+1)=8$ divisors and $1000$ has $(3+1)(3+1)=16$ divisors. As $n\ne 1$, the number of possible values for $n$ is $(8-1)+(16-1)=22$.