How many numbers in between $1$ and $250$ are not divisible by $2, 3, 5$ or $7$? How many numbers in between $1$ and $250$ are not divisible by $2, 3, 5$ or $7$? Using the Inclusion Exclusion Principle, I got the answer to be $57$ (i.e. $250-193$), but I'm worried I might have screwed up the arithmetic. Is there a quick and easy way to check whether long calculations like these are right?
 A: I will first consider all integers up to $210$, as it is the L.C.M. of $2$, $3$, $5$ and $7$. The number of integers from $1$ to $210$ which are not divisible by  $2$, $3$, $5$ or $7$ is
$$210-\frac{210}{2}-\frac{210}{3}-\frac{210}{5}-\frac{210}{7}+\frac{210}{6}+\frac{210}{10}+\frac{210}{14}+\frac{210}{15}+\frac{210}{21}+\frac{210}{35}-\frac{210}{30}-\frac{210}{42}-\frac{210}{70}-\frac{210}{105}+\frac{210}{210}$$
which is equal to
$$210\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)=48$$
The integers from $211$ to $250$ can be replaced by $1$ to $40$. I prefer to simply list out all integers satisfying the conditions: $1$, $11$, $13$, $17$, $19$, $23$, $29$, $31$, $37$.
So, there are $48+9=57$ such integers.
A: |A∪B∪C∪D|=|A|+|B|+|C|+|D|} all singletons−(|A∩B|+|A∩C|+|A∩D|+|B∩C|+|B∩D|+|C∩D|)} all pairs+(|A∩B∩C|+|A∩B∩D|+|A∩C∩D|+|B∩C∩D|)} all triples−|A∩B∩C∩D|} all quadruples
|A|+|B|+|C|+|D|=250/2 + 250/3 + 250/5 + 250/7 = 294
|A∩B|+|A∩C|+|A∩D|+|B∩C|+|B∩D|+|C∩D|= 250/(2*3) + 250/(2*5) + 250/(2*7) + 250/(3*5) + 250/(3*7) + 250/(5*7) = 120
|A∩B∩C|+|A∩B∩D|+|A∩C∩D|+|B∩C∩D|= 250/(2*3*5) + 250/(2*3*7) + 250/(2*5*7) + 250/(3*5*7) = 20 
|A∩B∩C∩D| = 1
|A∪B∪C∪D|=|A|+|B|+|C|+|D| − (|A∩B|+|A∩C|+|A∩D|+|B∩C|+|B∩D|+|C∩D|)  +(|A∩B∩C|+|A∩B∩D|+|A∩C∩D|+|B∩C∩D|) − |A∩B∩C∩D|
|A∪B∪C∪D|= 294 - 120 + 20 - 1 = 193
so no not divisible = 250 - 193 = 57 integers
