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How many natural-number factors does $N$ have if $N = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7^2$?

 Feb 17, 2021
 #1
avatar+268 
+1

We have a prime factorization of $2^4 \times 3^3 \times 5^2 \times 7^2$

If we add one to each exponent, then add up all the exponents.

$5 + 4 + 3 + 3 = 15$

 

Therefore there are 15 natural factors.

 

Hope that helps!!!

🙃🙃👌

 Feb 17, 2021
 #2
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0

Sorry, but that is wrong.

 Feb 17, 2021
 #3
avatar+120129 
+1

OBJD  is close....but....we  need  to add 1 to each exponent and then take that product....so..

 

(4+1)  (3+1) (2 + 1) ( 2 + 1)   =

 

5  *  4    *   3    *    3       =

 

180     factors

 

 

cool cool cool

 Feb 17, 2021

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