Collection of Maths Problems

Numbers not divisible by 6, 10 or 14

Task number: 3487

• 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: \(840 - 228 = 612\).

• Answer

  There are 612 such numbers in the given range.