How many numbers between $1 - 1000$ leave no remainder when divided by $4$ and when divided by $6$ but not when divided by $21$?

Question

Numbers between $1 - 1000$ which leave no remainder when divided by $4$ and divided by $6$ but not by $21$?

I tried $$\frac{1000}{12} = 83 - \frac{83}{21} = 83-3 = 80$$

Am I correct? Can someone please explain to me how it works?

Answer 1

If a number is divisible by both $4$ and $6$, then it is divisible by $\operatorname{lcm}(4, 6) = 12$. The number of multiples of $12$ that are at most $1000$ is $$\left\lfloor \frac{1000}{12} \right\rfloor = 83$$ where $\lfloor x \rfloor$ is the greatest integer less than $x$.

From these, we must subtract those numbers that are also divisible by $21$. Those numbers are divisible by $\operatorname{lcm}(4, 6, 21) = \operatorname{lcm}(12, 21) = 84$. The number of multiples of $84$ that are at most $1000$ is $$\left\lfloor \frac{1000}{84} \right\rfloor = 11$$ Hence, the number of positive integers less than or equal to $1000$ that are divisible by both $4$ and $6$ but not divisible by $21$ is $$\left\lfloor \frac{1000}{12} \right\rfloor - \left\lfloor \frac{1000}{84} \right\rfloor = 83 - 11 = 72$$

Answer 2

We get that the number is a multiple of 12 but not of 21. As 12 is coprime to 21, and $12 \times 21 = 252$, which has 3 multiples in 1-1000, we also know that there are 83 multiples of 12 in 1-1000. So, we get 83-3, which is equal to 3.