(a)$n=2p_k$, since $k>p_k$,$p_k>2p_k-k=n-k$. Since $p_k$ is a prime number,$p_k$ does not divide $(n-k)!$,$n$ does not divide $(n-k)!$. (b)$n>2p_k$, since $n$ is not prime,suppose $n=ab(2\leq a\leq b)$. (I)$a\geq3$,$n>2p_k\geq\frac{3k}{2}$,$\frac{2}{3}n>k,n-k>\frac{n}{3}$. (i)$a\neq b$.By $b\leq\frac{n}{3}$,$1<a<b<n-k$,$n=ab|(n-k)!$. (ii)$a=b$,$n=a^2,n-k>\frac{n}{3}=\frac{a^2}{3}$.By $k\geq14,p_k\geq13,n>26,a\geq6$. $\frac{a^2}{3}\geq2a,n-k>2a$,$\therefore a\cdot 2a|(n-k)^2,n|(n-k)^2$. (II)$a=2$,since $n>26$, (i)$b$ isn't prime,$b=b_1b_2(2\leq b_1\leq b_2)$. Since $b>13,b_2\geq4,ab_1\geq4,n=(ab_1)(b_2)$,both of $ab_1,b_2$ are larger than 3, so by (I),$n|(n-k)!$. (ii)$b$ is prime,$b=\frac{n}{2}>p_k$,since $p_k$be the largest prime number which is strictly less than k, $b\geq k$,$n-k=2b-k\geq b$,so $b|(n-k)!$,$n|(n-k)!$. So if $n>2p_k$,$n|(n-k)!$.

(a) for any integer $i$ such that $0\leq i\leq 2p_k-k-1$ $\implies k\leq i+k\leq 2p_k-1$ $\implies p_k<i+k<2p_k$...(*) If $n|(n-k)!$ so does $p_k|(2p_k-k)!$ or $p_k|(2p_k-k-i)$ . But this is impposible by (*) (b) havent solved

Since k > p, we have p > 2p-k and hence p does not divide any of 1, 2, 3, ... , 2p-k. But p is prime, so it does not divide (2p - k)! So 2p does not either. Now consider composite n > 2p. Note that k > 14, so p ≥ 13, so n > 26. Take q to be the largest prime divisor of n and put n = qr. We now have three cases. (1) q > r ≥ 3. Then n > 2p ≥ 3k/2, so 2n/3 > k. Hence n-k > n-2n/3 = n/3 ≥ n/r = q > r. So q and r are distinct integers < n-k. Hence n = qr divides (n-k)!. (2) r = 2. Then n = 2q > 2p. But p is the largest prime < k. Hence q ≥ k, so n-k ≥ n-q = q. Obviously q > 2 (since n > 26), so q and 2 are distinct integers < n-k. Hence n = 2q divides (n-k)! . (3) The final case is n = q2. We show that 2q ≤ n-k. Suppose not. Then 2q ≥ n-k+1 ≥ (2p+1)-k+1 ≥ 3k/2 + 1 - k + 1 = k/2 + 2. So q ≥ k/4 + 1. Also since 2q ≥ n-k+1, we have k ≥ n-2q+1 = (q-1)2 ≥ k2/16. If k > 16, this is a contradiction. If k = 15 or 16, then p = 13 and k ≥ (q-1)2 gives q ≤ 5, so n = q2 ≤ 25. But n > 2p = 26. Contradiction. So we have established that 2q ≤ n-k. But that means that q and 2q are distinct integers ≤ n-k and so their product 2n divides (n-k)!.

I got this, but I forgot the case where $n=q^2$ for part b. How many do you think I would get if ever? Thanks!

jin wrote: (a)$n=2p_k$, since $k>p_k$,$p_k>2p_k-k=n-k$. Since $p_k$ is a prime number,$p_k$ does not divide $(n-k)!$,$n$ does not divide $(n-k)!$. (b)$n>2p_k$, since $n$ is not prime,suppose $n=ab(2\leq a\leq b)$. (I)$a\geq3$,$n>2p_k\geq\frac{3k}{2}$,$\frac{2}{3}n>k,n-k>\frac{n}{3}$. (i)$a\neq b$.By $b\leq\frac{n}{3}$,$1<a<b<n-k$,$n=ab|(n-k)!$. (ii)$a=b$,$n=a^2,n-k>\frac{n}{3}=\frac{a^2}{3}$.By $k\geq14,p_k\geq13,n>26,a\geq6$. $\frac{a^2}{3}\geq2a,n-k>2a$,$\therefore a\cdot 2a|(n-k)^2,n|(n-k)^2$. (II)$a=2$,since $n>26$, (i)$b$ isn't prime,$b=b_1b_2(2\leq b_1\leq b_2)$. Since $b>13,b_2\geq4,ab_1\geq4,n=(ab_1)(b_2)$,both of $ab_1,b_2$ are larger than 3, so by (I),$n|(n-k)!$. (ii)$b$ is prime,$b=\frac{n}{2}>p_k$,since $p_k$be the largest prime number which is strictly less than k, $b\geq k$,$n-k=2b-k\geq b$,so $b|(n-k)!$,$n|(n-k)!$. So if $n>2p_k$,$n|(n-k)!$. I like your solution, but I think you meant to write $ a\cdot 2a|(n-k)!,n|(n-k)!$ instead of $a\cdot 2a|(n-k)^2,n|(n-k)^2$ on line seven.

Bashy problem shobber wrote: Let $k\ge 14$ be an integer, and let $p_k$ be the largest prime number which is strictly less than $k$. You may assume that $p_k\ge 3k/4$. Let $n$ be a composite integer. Prove: (a) if $n=2p_k$, then $n$ does not divide $(n-k)!$; (b) if $n>2p_k$, then $n$ divides $(n-k)!$. Solution: (a) We have \[n - k < 2p_k - p_k < p_k\]so $p_k \nmid (n-k)!$. (b) Suppose there is a prime $q$ such that $q\mid n$, but $q\nmid (n-k)!$. Therefore $q<n$ and $q>n-k$. If $n = 2q$, then $n=2q>2k$. Therefore $2q > 2n -2k > n = 2q$ which is a contradiction. If $n \neq 2q$, then as $n$ is composite, $n\geq 3q$. Therefore $n\geq 3q>3(n-k)\implies k>\frac{2n}{3}$ which is a contradiction. Therefore there is no such prime $q$. Suppose there is a prime $q$ such that $a=v_q(n)>v_q((n-k)!)$. Also, $a \geq 2$. We have $n\geq q^a$ and \[a>\lfloor \frac{q^{a-1}}{3} \rfloor > \frac{q^{a-1}}{3}-1\implies 3a+3> q^{a-1}.\]As $a\geq 2$, we need to check for primes $q=2,3,5,7$. A simple case bash for $q=2,3,5,7$ using the bounds we are done $~~\square$

Part A: Since $2p_k-k<p_k$ as $p_k<k$, $p_k$ does not divide $(2p_k-k)!$, so $2p_k$ also does not divide it. Part B: Case 1: $n$ can be expressed as $ab$ for $a,b\geq 3$. Note that this means $a,b\leq n/3$. Then, $$k\leq \frac{4}{3}p_k<\frac{2}{3}n,$$so $$n-k\geq \frac{1}{3}n.$$Thus, the factorial will be divisible by $ab$. Case 2: $n$ cannot be expressed in such a way. Then, we must have $n=2p$ for a prime $p$. Then, we wish to show $$2p\mid (2p-k)!.$$However, $p>p_k$, and since by definition $p_k$ is the largest prime under $k$, we must have $p\geq k$, so $2p-k\geq p.$ Thus, the factorial contains the factor of $p$ (and obviously a factor of 2 as well), so we are done.