If x, y, and z are positive integers such that 6xyz+30xy+21xz+2yz+105x+10y+7z=812, find x+y+z.
I'm skeptical about this problem.
Where did you get it?
Maybe some method to solve this, but I certainly do not know what it might be....
WolframAlpha gives (x, y, z) = (2, 2, 6)
Looks like it was posted here: https://www.reddit.com/r/cheatatmathhomework/comments/weka3/6xyz30xy21xz2yz105x10y7z_812_find_x_y_z/
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 .....!!!!
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.
+129999 
