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How many positive divisors of 8400 have at least 4 positive divisors?

 Mar 21, 2019
edited by Badada  Mar 21, 2019
 #1
avatar+102791 
+2

factor(840) = ((2^3*3)*5)*7 = (2^3) *3*5*7

so there are 6 prime factors (2 occurs 3 times)

 

 

 

I can have 3,5,7 and some of the 2s   (1,2,or 3 twos)           

3 ways

 

I can have two of 3,5,7 and 2 or more number of 2s  (2 or 3 twos)

3C2*2 = 6 ways

 

I can have one of 3,5,7 and  all  3 twos

3*1 = 3 ways

 

 

 

3+6+3 = 12 ways

 

So I think there are 12 factors of  840 that have at least 4 factors.

 Mar 21, 2019
edited by Melody  Mar 21, 2019
 #3
avatar+322 
+2

Melody you forgot a ten in your prime factorization, but it is fine i get it now. Thanks!

 Mar 21, 2019
 #4
avatar+102791 
0

10 is not prime.

 

I just got the site calculator to give me the factors.  I didn't check.

 

3*5*7*2*2*2 = 840    

 

Woops it was 8400 not 840    blush

 

Ok I am glad you caught my copy error!      wink

Melody  Mar 21, 2019
edited by Melody  Mar 21, 2019

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