Official Solution:If \(x\) and \(y\) are positive integers and \(xy - x! = 0\), where \(y < 100\), in how many pairs \((x, y)\) satisfying these conditions is \(x ≥ y\)? A. 1
B. 2
C. 3
D. 4
E. 5
Re-arrange to get \(xy = x!\).
Given that x! equals \((x - 1)! * x\), we have: \(xy = (x - 1)! * x\).
Reduce by x to get: \(y = (x - 1)!\).
Since \(y < 100\), we need to find such values of x where \((x - 1)!\) is less than 100.
The above condition is true for 5 values of \(x\): 1, 2, 3, 4, and 5.
If \(x = 1\), then \(y = (x - 1)! = 0! = 1\). In this case, \(x ≥ y\) is satisfied.
If \(x = 2\), then \(y = (x - 1)! = 1! = 1\). In this case, \(x ≥ y\) is satisfied.
If \(x = 3\), then \(y = (x - 1)! = 2! = 2\). In this case, \(x ≥ y\) is satisfied.
If \(x = 4\), then \(y = (x - 1)! = 3! = 6\). In this case, \(x ≥ y\) is NOT satisfied.
If \(x = 5\), then \(y = (x - 1)! = 4! = 24\). In this case, \(x ≥ y\) is NOT satisfied.
Therefore, out of 5 pairs of \((x, y)\), 3 pairs satisfy \(x ≥ y\): (1, 1), (2, 1), and (3, 2).
Answer: C