\(n! = 2^{15} \times 3^6 \times 5^3 \times 7^2 \times 11 \times 13\)
Find n.
I started this problem by noting that there has to be at least a 13.
1*2*3*4*5*6*7*8*9*10*11*12*13
This only has 2 5's so we go up to 15.
1*2*3*4*5*6*7*8*9*10*11*12*13*14*15
Not enough 2's so we use trial and error to find
1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16= 16!
Because there is a 72, we know that it has to be at least 14! and not as big as 21!
Starting at the bottom"
2! = 21
3! = 2131
4! = 2331
5! = 233151
6! = 243251
7! = 24325171
8! = 27325171
9! = 27345171
10! = 28345271
11! = 28345271111
12! = 210355271111
13! = 210355271111131
14! = 211355272111131
15! = 211365372111131
16! = 215365372111131