Number from 1 to 70 divisible by none of 2, 5, 7

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An integer is chosen uniformly at random from 1 to 70. What is the probability that it is divisible by none of 2, 5, and 7?

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Count "divisible by at least one" using inclusion-exclusion, then take the complement. Among 1 to 70: - singles: multiples of 2 give 35, of 5 give 14, of 7 give 10, - pairs: multiples of 10 give 7, of 14 give 5, of 35 give 2, - triple: multiples of 70 give 1. Then \(\text{at least one}) = (35 + 14 + 10) - (7 + 5 + 2) + 1 = 59 - 14 + 1 = 46. So 70 - 46 = 24 numbers avoid all three, and the probability is \frac{24}{70} = \frac{12}{35} \approx 0.3429. Since 70 = 2 \times 5 \times 7, this matches \tfrac12 \cdot \tfrac45 \cdot \tfrac67 = \tfrac{24}{70}, confirming the count. Takeaway. With coprime…

Number from 1 to 70 divisible by none of 2, 5, 7

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