Question #279375

How many positive integers less than 100 is not a factor of 2,3 and 5?

Expert's answer

Let A,BA,B and CC denote the sets of positive integers that are divisible by 2, 3 and 5 respectively and less or equal to 100.

Then ∣A∣=⌊1002⌋=50,∣B∣=⌊1003⌋=33,|A|=\lfloor \frac{100}2\rfloor=50, |B|=\lfloor \frac{100}3\rfloor=33, and ∣C∣=⌊1005⌋=20.|C|=\lfloor \frac{100}5\rfloor=20.

If mm and nn are relatively prime then a aa is divisible by mm and nn iff it is divisible by mn.mn.

It follows that ∣A∩B∣=⌊1006⌋=16,∣A∩C∣=⌊10010⌋=10,∣B∩C∣=⌊10015⌋=6,|A\cap B|=\lfloor \frac{100}6\rfloor=16, |A\cap C|=\lfloor \frac{100}{10}\rfloor=10, |B\cap C|=\lfloor \frac{100}{15}\rfloor=6, and ∣A∩B∩C∣=⌊10030⌋=3.|A\cap B\cap C|=\lfloor \frac{100}{30}\rfloor=3.

Therefore, by Inclusion-exclusion principle, the number of positive integers that are divisible by 2 or 3 or 5 and less or equal to 100 is equal to

∣A∪B∪C∣=∣A∣+∣B∣+∣C∣−∣A∩B∣−∣A∩C∣−∣B∩C∣+∣A∩B∩C∣=50+33+20−16−10−6+3=74.|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C| \\=50+33+20-16-10-6+3=74.

We conclude that the number of integers that are not divisible by 2 or 3 or 5 is equal to 100−74=26.100-74=26.

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