How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

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




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


Subscribe to Cheenta at Youtube

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.