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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

But try the problem first…

Source

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.