1976 AHSME Problems/Problem 30

Revision as of 14:42, 7 May 2020 by Someonenumber011 (talk | contribs) (Solution)

Problem 30

How many distinct ordered triples $(x,y,z)$ satisfy the equations \[x+2y+4z=12\] \[xy+4yz+2xz=22\] \[xyz=6\]

$\textbf{(A) }\text{none}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6$


Noticing the variables and them being multiplied together, we try to find a good factorization. After trying a few, we stumble upon something in the form of \[(x+*)(y+*)(z+*)\] where the blanks should be filled in with numbers. Filling in as \[(x+4)(y+2)(z+1)\] gives \[2x+4y+8z+xy+4yz+2xz+xyz+8\], and all parts happen to be multiples of the given equations. After substitution, we get \[(x+4)(y+2)(z+1)=60\].

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