: 6! = 8x9x10. What is the largest integer n for which n! is equal to the product of 3 consecutive integers?

Question

: 6! = 8x9x10. What is the largest integer n for which n! is equal to the product of 3 consecutive integers?

Answer 1

smallest integer n, n!= 3 consecutive integers is n=3

3!=3x2x1= 6

4!=4x3x2x1 = 2x3x4 = 24

6!=6x5x4x3x2 =720 = 8x9x10=720

but from 7! to 9!, the 3 consecutive integers have to have a product as a multiple of 10. That's only possible if the 3 integers' start with 8,9,10, 3,4, or 5. There are no such integers whose product = 7!,8! or 9!

closest is 16x17x18 = 4896 with 7!=5040

with n! n>9, the product of 3 consecutive integers has to be a multiple of 100. That's not possible. put any integer in the tens' digit combined with 8,9, 0, 3,4, or 5 and the product will never be a multiple of 100

6! is the largest n! that = a product of 3 consecutive integers