Problem
NT-B2-M05-P017 Consecutive Integers
#17
★★★★★ Level 5 of 5
Prove that the product of any \(n\) consecutive integers is divisible by \(n!\).
Prove \(v_p(A)\ge v_p(n!)\) for every prime \(p\).
Let \(A=(m+1)(m+2)\cdots(m+n)\). Among these \(n\) numbers, at least \(\lfloor n/p\rfloor\) are divisible by \(p\), at least \(\lfloor n/p^2\rfloor\) are divisible by \(p^2\), and so on. Therefore \(v_p(A)\ge\lfloor n/p\rfloor+\lfloor n/p^2\rfloor+\cdots=v_p(n!)\). This holds for every prime \(p\), so \(n!\mid A\).
Olympiad proof through all primes at once.