+0  
 
+2
484
2
avatar+140 

Three positive integers are each greater than $1$, have a product of $ 27000 $, and are pairwise relatively prime. What is their sum?

off-topic
 Mar 19, 2019
 #1
avatar+235 
+1

\(8 \cdot 27 \cdot 125 = 27000\), and they are all relatively prime. So the sum is \(8 + 27 + 125 = 160\). I got this by looking up 3 factors of 27000, and it gave me \(2^3 \cdot 3^3 \cdot 5^3 = 27000\).

 

Hope this helps!

 Mar 19, 2019
 #2
avatar+140 
+1

thanks

CorbellaB.15  Mar 19, 2019

8 Online Users