 Solution

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

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

Because

p \geq k + 2

, neither

(k + 1)!

nor

k!

contains a factor

p

. On the other hand,

p \leq 2 k + 1

, so

(2 k + 1)!

contains a factor

p

; and

2 p \geq 2 k + 4 > 2 k + 1,

so it contains no second factor

p

. Therefore

v_{p} ((k + 1 2 k + 1)) = v_{p} ((2 k + 1)!) - v_{p} ((k + 1)!) - v_{p} (k!) = 1.

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