A number divisible by 2, 3, or 5

Question

Draw an integer uniformly at random from \{1, 2, 3, \dots, 1000\}. What is the probability it is divisible by 2, by 3, or by 5 (at least one of them)?

Accepted Answer

Count multiples with inclusion–exclusion. Use \lfloor 1000/d \rfloor for the count of multiples of d up to 1000. Singles: \lfloor 1000/2\rfloor = 500,\quad \lfloor 1000/3\rfloor = 333,\quad \lfloor 1000/5\rfloor = 200. Pairs (multiples of the products, since 2, 3, 5 are pairwise coprime): \lfloor 1000/6\rfloor = 166,\quad \lfloor 1000/10\rfloor = 100,\quad \lfloor 1000/15\rfloor = 66. Triple: \lfloor 1000/30\rfloor = 33. Combine: (500+333+200) - (166+100+66) + 33 = 1033 - 332 + 33 = 734. So 734 of the 1000 numbers qualify, giving P = \frac{734}{1000} = 0.734. Takeaway. For "divisible by at le…

A number divisible by 2, 3, or 5

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