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

 Apr 30, 2020
 #1
avatar+50 
0

Stop posting AoPs questions...

Don't use this as a cheat site.

 

 

NO ONE ANSWER THIS QUESTION...

 Apr 30, 2020
 #2
avatar+657 
+1

Yes I agree unless the person who asks a question has a direct question about a subject

LuckyDucky  Apr 30, 2020

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