 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/11g/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Every prime in

(X, X]

is at least

X

. Consequently the primorial satisfies

P (X) \geq (X)^{π (X) - π (X)} .

Combining this with part (c) and taking logarithms gives

(π (X) - π (X)) \frac{l o g X}{2} \leq X lo g 4,

π (X) \leq π (X) + \frac{2 X l o g 4}{l o g X} \leq X + \frac{2 X l o g 4}{l o g X} .

For

X \geq 2

, the inequality

lo g X \leq X

implies

X \leq \frac{X}{l o g X} .

π (X) \leq (1 + 2 lo g 4) \frac{X}{l o g X} .

We may take

c = 1 + 2 lo g 4

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