Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on consecutive positive integers.
Someone observed that 6!=(8)(9)(10). Find the largest positive integer n for which n! can be expressed as the product of n-3 consecutive positive integers.
Integers
Inequality
Algebra
Answer: is 23.
AIME I, 1990, Question 11
Elementary Number Theory by David Burton
The product of (n-3) consecutive integers=\(\frac{(n-3+a)!}{a!}\) for a is an integer
\(n!=\frac{(n-3+a)!}{a!}\) for \(a \geq 3\) \((n-3+a)! \geq n!\)
or, \(n!=\frac{n!(n+1)(n+2)....(n-3+a)}{a!}\)
for a=4, n+1=4! or, n=23 which is greatest here
n=23.
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