Subjects combinatorics

Count Non Multiples B2E915

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Count Non Multiples B2E915


1. Problem: Find how many natural numbers $\leq 1000$ are not multiples of 4, 5, or 6. 2. Use the Inclusion-Exclusion Principle: - Total numbers from 1 to 1000: $1000$ - Multiples of 4: $\left\lfloor \frac{1000}{4} \right\rfloor = 250$ - Multiples of 5: $\left\lfloor \frac{1000}{5} \right\rfloor = 200$ - Multiples of 6: $\left\lfloor \frac{1000}{6} \right\rfloor = 166$ 3. Find multiples of pairwise least common multiples (LCM): - LCM(4,5) = 20, multiples: $\left\lfloor \frac{1000}{20} \right\rfloor = 50$ - LCM(4,6) = 12, multiples: $\left\lfloor \frac{1000}{12} \right\rfloor = 83$ - LCM(5,6) = 30, multiples: $\left\lfloor \frac{1000}{30} \right\rfloor = 33$ 4. Find multiples of LCM(4,5,6): - LCM(4,5,6) = 60, multiples: $\left\lfloor \frac{1000}{60} \right\rfloor = 16$ 5. Apply Inclusion-Exclusion: $$\text{Count} = 1000 - (250 + 200 + 166) + (50 + 83 + 33) - 16$$ Calculate stepwise: - Sum of singles: $250 + 200 + 166 = 616$ - Sum of pairs: $50 + 83 + 33 = 166$ So, $$\text{Count} = 1000 - 616 + 166 - 16 = 1000 - 616 + 150 = 534$$ 6. Therefore, there are $\boxed{534}$ natural numbers $\leq 1000$ that are not multiples of 4, 5, or 6.