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

 

Find n.

 Jun 30, 2020
 #1
avatar+297 
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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
avatar+21568 
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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

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