 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/2/a/solution

6, 8, 11

{1, \dots, 527}

24, 66, 88

264

. Therefore the count is

⌊ \frac{527}{6} ⌋ + ⌊ \frac{527}{8} ⌋ + ⌊ \frac{527}{11} ⌋ - ⌊ \frac{527}{24} ⌋ - ⌊ \frac{527}{66} ⌋ - ⌊ \frac{527}{88} ⌋ + ⌊ \frac{527}{264} ⌋ = 87 + 65 + 47 - 21 - 7 - 5 + 2 = 167 .

The endpoint

528

is excluded, which is why the numerators are

527

rather than

528

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