4

Part of 2013 IMAC Arhimede

Problems(1)

gcd ( [\frac{n}{p} ], (p-1)! ) = 1

Source: IMAC Arhimede 2013 p4

p,n

be positive integers, such that

p

p <n

p

n + 1

\left(\left[\frac{n}{p}\right], (p-1)!\right) = 1

, then prove that

p\cdot \left[\frac{n}{p}\right]^2

{n \choose p} -\left[\frac{n}{p}\right]

[x]

represents the integer part of the real number

x

number theorygreatest common divisorfloor function