If x is a positive integer, what is the value of x ?

Official Solution:

If \(x\) is a positive integer, what is the value of \(x\)?

The question does not need rephrasing, although we should note that \(x\) is a positive integer.

Statement 1: SUFFICIENT. We should work from the inside out by first listing the first several values of \(x!\) (the factorial of \(x\), defined as the product of all the positive integers up to and including \(x\)).
\(1! = 1\)
\(2! = 2\)
\(3! = 6\)
\(4! = 24\)
\(5! = 120\)
\(6! = 720\)
\(7! = 5040\)

Now we consider decimal expansions whose first nonzero digit is in the hundredths place. Such decimals must be smaller than \(0.1\) (\(\frac{1}{10}\)) but at least as large as \(0.01\) (\(\frac{1}{100}\)). Therefore, for \(\frac{1}{x!}\) to lie in this range, \(x!\) must be larger than 10 but no larger than 100. The only factorial that falls between 10 and 100 is \(4! = 24\), so \(x = 4\).

(Note that factorials are akin to exponents in the order of operations, so \(\frac{1}{x!}\) indicates "1 divided by the factorial of \(x\)," not "the factorial of \(\frac{1}{x}\)," which would only have meaning if \(\frac{1}{x}\) were a positive integer.)

Statement 2: INSUFFICIENT. We consider decimal expansions whose first nonzero digit is in the thousandths place. Such decimals must be smaller than \(0.01\) (\(\frac{1}{100}\)) but at least as large as \(0.001\) (\(\frac{1}{1000}\)). Therefore, for \(\frac{1}{(x+1)!}\) to lie in this range, \((x+1)!\) must be larger than 100 but no larger than 1,000.

There are two factorials that fall between 100 and 1,000, namely \(5! = 120\) and \(6! = 720\). Thus, \(x+1\) could be either 5 or 6, and \(x\) could be either 4 or 5.

Answer: A.