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Suppose that x,y,z are positive integers satisfying x ≤ y ≤ z, and such that the product of all three numbers is twice their sum. What is the sum of all possible values of z?

 Jul 2, 2018
 #1
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xyz =2[x+y+z], solve for x

 

x = (2 (y + z))/(y z - 2) and yz cannot=2

Therefore x cannot be > 1, because it will violate x ≤ y ≤ z

I can only see 2 solutions as follows:

x = 1,  y = 3   z = 8, and

x = 1,  y = 4,  z = 5

Sum of z =5 + 8 = 13

 Jul 2, 2018
 #2
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I will admit that I am tired but you lost me on the third line.

Melody  Jul 2, 2018
 #3
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It's 17

 

you frogot 2,2,4

matth99  Jul 3, 2018

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