Number from 1 to 100 divisible by none of 2, 3, 5

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An integer is chosen uniformly at random from 1 to 100. What is the probability that it is divisible by none of 2, 3, and 5?

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Because 100 is not a multiple of 30 = 2 \times 3 \times 5, we cannot rely on a clean period and must count carefully. Use inclusion-exclusion, rounding each count down. Among 1 to 100: - singles: multiples of 2 give 50, of 3 give 33, of 5 give 20, - pairs: multiples of 6 give 16, of 10 give 10, of 15 give 6, - triple: multiples of 30 give 3. Then \(\text{at least one}) = (50 + 33 + 20) - (16 + 10 + 6) + 3 = 103 - 32 + 3 = 74. So 100 - 74 = 26 numbers avoid all three, and the probability is \frac{26}{100} = \frac{13}{50} = 0.26. This is close to but not exactly the product \tfrac12 \cdot \tfra…

Number from 1 to 100 divisible by none of 2, 3, 5

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