Distributive factorials

Solution 1: When $k = 1$ this is clearly satisfied for all $n$ , since the equation simplifies to $n! = n!$ . When $n = k$ , the equation is again satisfied. The right hand side will be a product of $n - k$ consecutive integers, the largest of which is $n$ . The left hand side will be a product of $\frac{n}{k}$ consecutive integers, the largest of which is $\frac{n}{k}$ . When $n - k > \frac{n}{k}$ , this equation can never be equal. This inequality can be rearranged as $n >k + 1 + \frac{1}{k-1}, k \neq 1$ . Since $k \mid n$ , $n \geq k$ , this leaves $n = k+1$ and $k = 2, n = 4$ as the only possibilities not excluded by this, or already considered. When $n = k + 1$ and $k \mid n$ , we have $k \mid k + 1$ , so $k \mid 1$ and thus $k = 1$ . We have already considered this case. When $k = 2, n = 4$ , we have $2 = \left(\frac{4}{2}\right)! \neq \frac{4!}{2!} = 12$ .

Thus we see that the only possible solutions are when $k = 1$ or $n = k$ , by the principle of inclusion and exclusion, there are $50 + 50 - 1 = 99$ such solutions.

Solution 2: Here is another way to justify that $k$ must be $1$ or $n$ . Suppose $n=km.$ Then
$k!m!=(1\cdot 2 \cdot ...\cdot k)\cdot (1\cdot 2 \cdot...\cdot m ) =\frac{1}{k^{m-1}}\left(1\cdot ...\cdot(k-1)\right)\cdot \left((k\cdot 1)\cdot ...\cdot(k\cdot m)\right)\leq \frac{1}{k^{m-1}}(km)!$ So if both $k$ and $m$ are greater than $1$ , $k!m!<(km)!$

The equation can only be true if either n=k or k=1. There are 50 pairs of values where n=k and 50 pairs of values where k=1. However, n=1, k=1 satisfies both n=k and k=1 and as such counted twice. So, using inclusion-exclusion principle total number of pairs = 50+50-1=99

for k=1,n can take 50 values from 1 to 50 which will satisfy our condition so 50 ordered pairs ; and 49 pairs of n=k such as(2,2),(3,3)........................(50,50) [not including(1,1) because it is included in the above case]

therefore total ordered pairs=99

Let x=n/k, x integer. We need an x such as x! = n!/k!. So, n!=k!x!. But n=kx. Then, we need an x and a k such as (kx)!=k!x!. That only happens when x=1 or k=1. If k=1, n can be from 1 to 50. If x=1, n=k and n can be from 1 to 50. We have 50 pairs on the first one and 50 on the second. But the solution (1,1) is counted two times. So, we there are 99 solutions.

as (n/k)! would be too small in comparison to n!/k! for distinct n and k.either k=1 or k=n can satisty the given relation.

Para que a igualdade seja satisfeita há apenas duas condições:

1) Que n=k; ou 2) Que k=1.

São 50 números, de 1 até 50, cada um deles apresenta-se em 2 das duas condições, com exceção do 1, onde ambas condições são idênticas. (Todo número divide por 1 e por ele mesmo, 1 é ele mesmo)

Portanto 2x50-1=99.

1) A igualdade é satisfeita se n=k, por motivos óbvios. (n/k)!=(n/n)!=1!=1 (1o membro) n!/k!=n!/n!=1 (2o membro)

2) A igualdade também é satisfeita se k=1, também por motivos óbvios: (n/1)!=n! n!/1!=n!/1=n

É fácil ver que em quaisquer outras condições a igualdade não é satisfeita: Se k for um divisor de n, (n/k)! é um fatorial de um certo número. Já n!/k! será o número n(n-1)....(k+1).1 essa expressão só é um fatorial se k+1=2 (colocamos o 1 propositalmente para compreensão da idéia) ou se n=k (basta ler o produto 'ao contrário')

The above holds true iff (k=1) || (n==k). The union of those sets has size 99 (n=1 and k=1 is counted twice). We can see this because no factorial can equal a greater factorial divided by a lesser factorial.

let n =kt => k!t!=(kt)! 1 2 3 4 ... k ((LHS.))= (t+1)(t+2)(t+3)...(kt) ((RHS)) >=2 3 4 ...*k

LHS.is multiple of the number k RHS.is multiple of the number kt-t So k >= kt-t <=> k+t >= kt

case 1) k > t ; k+k > k+t >= kt so 2 > t => t =1 is posible (and work!)

case 2) k < t ; so k = 1 (and work!)

case 3) k = t ; if t >2 Apparently LHS. < RHS. so k = t = 1 (and work! lol)

so (n,k)=(kt,k) is possible is n=k or k = 1 have 50 + 50 -1 =99 #

k must be 1 or n, proof by contradiction(suppose k is not 2 or n: (n/k)! is n! divided by the product of all integers between n and (n/k)+1. There are n-n/k such integers. Notice n/k is between 2 and n/2-1, so 1>n-n/k>=k. So we have a product of at least k integers, each larger than k. Hence this product is larger than k!.

From here (n/k)!<n!/k!

By Bertrand's Postulate, $\exists$ prime $p\in [n,2n]$ . Since $n!=\left(\frac{n}{k}\right)!k!$ we have $p|k$ or $p|\frac{n}{k}$ which means either $k>\frac{n}{2}$ or $\frac{n}{k}>\frac{n}{2}\implies k=1,n$ Thus $n=1\implies k=1$

$n=2\implies k=1,2$

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$n=50\implies k=1,50$

Thus $99$ ordered pairs in total.