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Consecutive positive Integers | AIME I, 1990| Question 11

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


Answer: is 23.

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.

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