If x, y, and z are positive integers, and (x!+x)/z=y, then

First note that for \(n>1\), \(n!+1\) has no factor \(k<=n\), but \(1\). As \(n!=2*3*...*n\), which means that every integer less than or equal to \(n\) is a factor of \(n!\), but if we add \(1\) to \(n!\), sum, \(n!+1\), obviously won't be evenly divisible by these factors.

For example:
\(n=3\), \(n!+1=7\), \(7\) has no factor \(<=3\), except \(1\).
\(n=4\), \(n!+1=25\), \(25\) has no factor \(<=4\), except \(1\).
\(n=5\), \(n!+1=121\), \(121\) has no factor \(<=5\), except \(1\).

Back to the question:

If \(x\), \(y\), and \(z\) are positive integers, and \(\frac{x!+x}{z}=y\), then what is the value of \(z\)?

(1) \(x\) is a factor of \(y\) --> \(xk=y\) --> \(x!+x=yz=xkz\) --> \((x-1)!+1=kz\) --> \(z=1\) OR \(z>x-1\). Not sufficient.

For example:
\(x=4\), \((x-1)!+1=7=kz\), \(z=1\) OR \(z=7>4-1=3\)
\(x=5\), \((x-1)!+1=25=kz\), \(z=1\) OR \(z=5>5-1=4\) OR \(z=25>5-1=4\)
\(x=6\), \((x-1)!+1=121=kz\), \(z=1\) OR \(z=11>5-1=4\) OR \(z=121>5-1=4\)

(2) \(z<x\). Clearly insufficient.

For example:
\(x=4\), \(x!+x=28=yz\), \(z=1\) OR \(z=2\) OR \(z=4\) OR \(z=14\) OR \(z=28\).

(1)+(2) \(z=1\) OR \(z>x-1\) AND \(z<x\) --> As \(z<x\) and \(z\) is an integer, \(z>x-1\) can not be true, hence \(z=1\).

Answer: C.