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If x, y, and z are positive integers such that 6xyz+30xy+21xz+2yz+105x+10y+7z=812, find x+y+z.

 Dec 27, 2018
 #1
avatar+4394 
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I'm skeptical about this problem.

 

Where did you get it?

 Dec 27, 2018
 #2
avatar+98005 
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Maybe some method to solve this, but I certainly do not know what it might be....

 

WolframAlpha  gives   (x, y, z)  =  (2, 2, 6)

 

 

cool cool cool

 Dec 27, 2018
 #3
avatar+3987 
+1

Looks like it was posted here: https://www.reddit.com/r/cheatatmathhomework/comments/weka3/6xyz30xy21xz2yz105x10y7z_812_find_x_y_z/

 Dec 27, 2018
 #4
avatar+98005 
+1

Thanks, tertre.....not sure how that person knew that adding 35 to both sides would produce a "workable" problem, but....it was a good insight  .....!!!!

 

 

cool cool cool

CPhill  Dec 27, 2018
 #5
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If you assume that the thing on the left hand side can be written as

(ax + b)(cy + d)(ez + f) where a,b,c,d,e,f are some unknown constants,

expand and equate coefficients then everything drops out. 

The 35 appears naturally from the expansion, no insight needed.

 Dec 28, 2018

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