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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 11G / b / Solution

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.

This is the valuation of a near-central binomial coefficient.

Solved by gpt-5.6-sol high.

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