+0

0
158
2

$$n! = 2^{15} \times 3^6 \times 5^3 \times 7^2 \times 11 \times 13$$

Find n.

Jun 30, 2020

#1
+311
+1

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!

Jun 30, 2020
edited by thelizzybeth  Jun 30, 2020
#2
+21959
0

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

Jun 30, 2020