How many numbers between 1 and 100 are divisible by 2 or 3 or 5 or 7?
The solution I had gives a different answer from what was provided, so I was wandering if anyone could tell me what mistake I made?
Let A be the set of those divisible by 2, B the set of those divisible by 3, C the set of those divisible by 5 and D the set of those divisible by 7.
Singles:
$|A| = 50$, $|B| = 33$, $|C| = 20$, $|D| = 14$
Doubles:
$|A \cap B|=16$, $|A \cap C|=10$,
$|A \cap D|=7$, $|B \cap C|=6$,
$|B \cap D|=4$, $|C \cap D|=2$
Triples:
$|A \cap B \cap C|=3$, $|A \cap B \cap D|=2$
$|A \cap C \cap D|=1$, $|B \cap C \cap D|=0$
Quadruples:
$|A \cap B \cap C \cap D|=0$
Inclusion Exclusion:
$|A \cup B \cup C \cup D| = $ Sum of Singles $-$ Sum of Doubles $+$ Sum of Triples $-$ Sum of Quadruples
$ |A\cup B\cup C\cup D|=|A|+|B|+|C|+|D|\\ -(|A\cap B|+|A\cap C|+|A\cap D|+|B\cap C|+|B\cap D|+|C\cap D|)\\ +(|A\cap B\cap C|+|A\cap B\cap D|+|A\cap C\cap D|+|B\cap C\cap D|)\\ -|A\cap B\cap C\cap D|\\ $
$|A \cup B \cup C \cup D| = 50 + 33 + 20 + 14 - 16 - 10 - 7- 6 - 4 - 2 + 3 + 2 + 1$
$|A \cup B \cup C \cup D| = 114 - 45 + 6$
$|A \cup B \cup C \cup D| = 75$
The answer given to us was 78, but I can't tell whether it's a mistake or if I've made a mistake somewhere here.
$$50+33+20+14=11\color{red}{7}.$$