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for every prime p there is an integer k for which p divides k and k is a peak

Source: Canada Repêchage 2019/8 CMOQR

March 2, 2020

number theorydividesinequalities

Problem Statement

For

t \ge 2

, define

S(t)

as the number of times

t

divides into

t!

. We say that a positive integer

t

is a peak if

S(t) > S(u)

for all values of

u < t

. Prove or disprove the following statement: For every prime

p

, there is an integer

k

for which

p

divides

k

and

k

is a peak.