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

Source: IMAC Arhimede 2013 p4

May 6, 2019
number theorygreatest common divisorfloor function

Problem Statement

Let p,np,n be positive integers, such that pp is prime and p<np <n. If pp divides n+1n + 1 and ([np],(p−1)!)=1 \left(\left[\frac{n}{p}\right], (p-1)!\right) = 1, then prove that p⋅[np]2p\cdot \left[\frac{n}{p}\right]^2 divides (np)−[np]{n \choose p} -\left[\frac{n}{p}\right] . (Here [x][x] represents the integer part of the real number xx.)
Back to ProblemsView on AoPS