If 9!/3^k is integer, what is the greatest possible value of K?
qa is c - 4
9
6
4
3
2
qa is c - 4
9
6
4
3
2
Appetite for 700 and I scraped my plate!
Essentially you are being asked how many 3's you have in 9!.
In order for 9! / 3^k to be an integer, you must have at least k 3's in the numerator.
9! has four 3's: the 9 accounts for two of them, and the 6 and 3 for one each.
