How many numbers between 1 and 900 cannot be divided by 4, 5 or 6

Question

I'm trying to find total amount of numbers between 1 and 900 that is not dividable by 4, 5 or 6. For example, number 16 is not part of the total because 16 is dividable with 4. I use inclusion and exclusion and was told the answer is 480. However, my calculations don't give the same result :(

$$ \left| A \cup B \cup C \right| = \left| A \right|+\left| B \right|+\left| C \right|-\left| A \cap B \right| - \left| A \cap C \right| - \left| B \cap C \right| + \left| A \cap B\cap C \right| = \frac{900}{4} + \frac{900}{5}+ \frac{900}{6} - \frac{900}{4\cdot 5} - \frac{900}{4\cdot 6} - \frac{900}{5\cdot 6} + \frac{900}{4\cdot 5 \cdot 6} =225+180+150-45-37.5-30+7.5=450 $$

Where am I doing wrong? Please help.

Answer 1

You should have noticed that those fractional quantities $37.5$, $7.5$ make no sense here! $900$ is divisible by $4,5,$ and $6$, so you don't have to worry about what happens to remainders. But you do have to use $\mathrm{lcm}(4,6)$ and $\mathrm{lcm}(4,5,6)$ in the denominators instead of $4\cdot 6$ and $4\cdot 5\cdot 6$.

Answer 2

Thank to TonyK answer the solution can be written as

$$ \left| A \cup B \cup C \right| = \left| A \right|+\left| B \right|+\left| C \right|-\left| A \cap B \right| - \left| A \cap C \right| - \left| B \cap C \right| + \left| A \cap B\cap C \right| = \frac{900}{4} + \frac{900}{5}+ \frac{900}{6} - \frac{900}{4\cdot 5} - \frac{900}{lcm(4,6)} - \frac{900}{5\cdot 6} + \frac{900}{lcm(4,5,6)} = \frac{900}{4} + \frac{900}{5}+ \frac{900}{6} - \frac{900}{4\cdot 5} - \frac{900}{12} - \frac{900}{5\cdot 6} + \frac{900}{60} = 225+180+150-45-75-30+15=420 $$ Hence the sought total is $900-420=480$.