Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on consecutive positive integers.

Consecutive positive integer – AIME I, 1990

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.

  • is 107
  • is 23
  • is 634
  • cannot be determined from the given information

Key Concepts




Check the Answer

But try the problem first…

Answer: is 23.

Suggested Reading

AIME I, 1990, Question 11

Elementary Number Theory by David Burton

Try with Hints

First hint

The product of (n-3) consecutive integers=\(\frac{(n-3+a)!}{a!}\) for a is an integer

Second Hint

\(n!=\frac{(n-3+a)!}{a!}\) for \(a \geq 3\) \((n-3+a)! \geq n!\)

or, \(n!=\frac{n!(n+1)(n+2)….(n-3+a)}{a!}\)

Final Step

for a=4, n+1=4! or, n=23 which is greatest here


Subscribe to Cheenta at Youtube