# Inclusion-Exclusion Principle application: numbers $\le1000$ which are multiples of $3$ or $7$ or $12$

I am suppose to find the number of integers from 1 to 1000 inclusive which are multiples of 3 or 7 or 12:

The answer I got was 414 however, the answer stated on the answer sheet provided is 428. Can anybody explain to me why is it 428?

My working is as shown:

$A=3,\:B=\:7,\:C=\:12$

$\left|A\right|=333\:,\:\left|B\right|=\:142,\:\left|C\right|=\:83$

$\left|A\bigcup B\bigcup C\right|=\left|A\right|+\left|B\right|+\left|C\right|-\left|A\bigcap B\right|-\left|B\bigcap C\right|-\left|A\bigcap C\right|-\left|A\bigcap B\bigcap C\right|\:=333\:+142\:+83\:-47-83-11-3\:=\:414$

Appreciate any help!

Since $3 \mid 12$, we know that every multiple of $12$ is also a multiple of $3$, so that $C \subseteq A$. So we're really only dealing with two sets: $$|A \cup B \cup C| = |A \cup B| = |A| + |B| - |A \cap B| = 333 + 142 - 47 = 428$$
Actually, all multiples of 12 are multiples of 3, so all you are looking for are numbers that are either multiples of 3 or multiples of 7. Thus, $\lfloor \frac{1000}{3} \rfloor + \lfloor \frac{1000}{7} \rfloor - \lfloor \frac{1000}{21} \rfloor = 333 + 142 - 47 = 428.$