There are 734 integers between 1 and 1000 inclusive that are evenly divisible by 2, 3, or 5.
Based on the provided reference, we know that there are 734 positive integers from 1 to 1000 that are divisible by at least one of the numbers 2, 3, or 5. This result is obtained using the principle of inclusion-exclusion, which is a common technique used to solve problems of this nature. The reference states, "Therefore, there are 734 positive integers which are divisible by at least 2, 3 or 5 from 1 to 1000."
Here's a breakdown of why the principle of inclusion-exclusion is used:
- Divisibility by 2: There are 1000 / 2 = 500 numbers divisible by 2.
- Divisibility by 3: There are 1000 / 3 ≈ 333 numbers divisible by 3.
- Divisibility by 5: There are 1000 / 5 = 200 numbers divisible by 5.
However, simply adding these would count numbers divisible by both 2 and 3 (i.e., divisible by 6), both 2 and 5 (i.e., divisible by 10), and both 3 and 5 (i.e., divisible by 15) twice. We would also over count the number divisible by all three: 2, 3 and 5 (i.e., divisible by 30). Therefore we need to subtract off those over counted intersections, and then re-add the intersection of all three.
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Divisibility by 6: There are 1000 / 6 ≈ 166 numbers divisible by 6.
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Divisibility by 10: There are 1000 / 10 = 100 numbers divisible by 10.
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Divisibility by 15: There are 1000 / 15 ≈ 66 numbers divisible by 15.
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Divisibility by 30: There are 1000 / 30 ≈ 33 numbers divisible by 30.
Using the principle of inclusion-exclusion:
Total = (Divisible by 2) + (Divisible by 3) + (Divisible by 5) - (Divisible by 6) - (Divisible by 10) - (Divisible by 15) + (Divisible by 30)
Total = 500 + 333 + 200 - 166 - 100 - 66 + 33 = 734
Therefore, there are 734 integers between 1 and 1000 (inclusive) that are divisible by 2, 3, or 5.
