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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?

 Jan 3, 2023
 #1
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 318 , 415 , 514 , 813>> Total = 4

 Jan 3, 2023
 #2
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The answer is 22.

Guest Jan 3, 2023
 #3
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22 is incorrect guest, but thanks for trying!

Keihaku  Jan 3, 2023
 #4
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4 is also incorrect, guest. Thanks for your effort! 

Keihaku  Jan 3, 2023
 #5
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1  = (2, 2, 4)
2  = (1, 4, 5)
3  = (1, 3, 8)

Guest Jan 3, 2023
 #6
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So would the answer be 

(2+2+4)+(1+4+5)+(1+3+8)?

Keihaku  Jan 3, 2023
 #7
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Thank you, but it says that is incorrect.

Keihaku  Jan 3, 2023
 #8
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1 - (2 * 2 * 4) ==2 * (2 + 2 + 4) ==16 = 16

 

2 - (1 * 4 * 5) ==2 * (1 + 4 + 5) ==20 = 20

 

3 - (1 * 3 * 8) ==2 * ( 1+ 3 + 8) ==24 = 24

Guest Jan 3, 2023
 #9
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Sum of all possible value of z ==4 + 5 + 8 ==17

Guest Jan 3, 2023
 #10
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Both of those are also incorrect, I'm so sorry. I appreciate your help though!

Keihaku  Jan 3, 2023
 #11
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159  167  235  333  Total =  4

 

 

Then the sum of all possible values of z is 9 + 7 + 5 + 3 = 24.

 Jan 4, 2023
 #12
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That was wrong, sorry.

Keihaku  Jan 4, 2023

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