Number from 1 to 105 divisible by none of 3, 5, 7

Question

An integer is chosen uniformly at random from 1 to 105. What is the probability that it is divisible by none of 3, 5, and 7?

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Count "divisible by at least one" using inclusion-exclusion, then take the complement. Among 1 to 105: - singles: multiples of 3 give 35, of 5 give 21, of 7 give 15, - pairs: multiples of 15 give 7, of 21 give 5, of 35 give 3, - triple: multiples of 105 give 1. Then \(\text{at least one}) = (35 + 21 + 15) - (7 + 5 + 3) + 1 = 71 - 15 + 1 = 57. So 105 - 57 = 48 numbers avoid all three, and the probability is \frac{48}{105} = \frac{16}{35} \approx 0.4571. Since 105 = 3 \times 5 \times 7, this also equals \left(1-\tfrac13\right)\left(1-\tfrac15\right)\left(1-\tfrac17\right) = \tfrac23 \cdot \tfra…

Number from 1 to 105 divisible by none of 3, 5, 7

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