# Difference between revisions of "1976 AHSME Problems/Problem 30"

## Problem 30

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

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

## Solution

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 $$2x+4y+8z+xy+4yz+2xz+xyz$$, and all parts happen to be multiples of the given equations. After substitution, we get $$(x+4)(y+2)(z+1)=52$$.