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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**

Integers

Inequality

Algebra

## Check the Answer

But try the problem first…

Answer: is 23.

Source

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

n=23.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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