Numbers not divisible by 6, 10 or 14

Determine the number of integers between 1 and 840 that are not divisible by 6, 10 nor 14.

• Solution

Let $$A_1$$ be the set of multiples of six from $$\{1,…,840\}$$, similarly let $$A_2$$ be the multiples of 10 and $$A_3$$ be the multiples of 14. We will use $$A_{1{,}2}$$ to denote the common multiples of 6 and 10, i.e. $$A_1\cap A_2$$, etc.

Then $$|A_1|=\left\lfloor \frac{840}{6}\right\rfloor=140$$, $$|A_2|=\left\lfloor \frac{840}{10}\right\rfloor=84$$, $$|A_3|=\left\lfloor \frac{840}{14}\right\rfloor=60$$, $$|A_{1{,}2}|=\left\lfloor \frac{840}{lcm(6{,}10)}\right\rfloor=\left\lfloor \frac{840}{30}\right\rfloor=28$$, $$|A_{1{,}3}|=\left\lfloor \frac{840}{lcm(6{,}14)}\right\rfloor=\left\lfloor \frac{840}{42}\right\rfloor=20$$, $$|A_{2{,}3}|=\left\lfloor \frac{840}{lcm(10{,}14)}\right\rfloor=\left\lfloor \frac{840}{70}\right\rfloor=12$$ and $$|A_{1{,}2,3}|=\left\lfloor \frac{840}{lcm(6{,}10,14)}\right\rfloor=\left\lfloor \frac{840}{210}\right\rfloor=4$$.

By the inclusion-exclusion principle, the total number of multiples of 6, 10 or 14 is $$|A_1\cup A_2\cup A_3|=$$$$|A_1|+|A_2|+|A_3|-|A_{1{,}2}|-|A_{1{,}3}|-|A_{2{,}3}|+|A_{1{,}2,3}| =140+84+60-28-20-12+4=228.$$

It remains to calculate the complement of this set: $$840 - 228 = 612$$.