MathLabs

Problem 3

Let k≥14k\ge14 be an integer, and let pkp_k be the largest prime strictly less than kk. You may assume that pk≥3k/4p_k\ge3k/4. Let nn be composite. Prove: (a) if n=2pkn=2p_k, then nn does not divide (n−k)!(n-k)!; (b) if n>2pkn>2p_k, then nn divides (n−k)!(n-k)!.
Step 1 of 5: Handle the boundary case n=2pk
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n=2pk  ⟹  n−k=2pk−k<pkn=2p_k\implies n-k=2p_k-k<p_k
Detailed analysis

Since pk<kp_k<k, we have n−k=2pk−k<2pk−pk=pkn-k=2p_k-k<2p_k-p_k=p_k. Thus (n−k)!(n-k)! contains no multiple of the prime pkp_k, so it is not divisible by pkp_k, and therefore cannot be divisible by n=2pkn=2p_k.