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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.

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


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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