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

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== Solution ==
 
== 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 <math>(x+-)(y+-)(z+-)</math> where the blanks should be filled in. Filling in as <math>(x+4)(y+2)(z+1)</math> gives <math>2x+4y+8z+xy+4yz+2xz+xyz</math>, and all parts happen to be multiples of the given equations.
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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 <math>(x+*)(y+*)(z+*(</math> where the blanks should be filled in with numbers. Filling in as <math>(x+4)(y+2)(z+1)</math> gives <math>2x+4y+8z+xy+4yz+2xz+xyz</math>, and all parts happen to be multiples of the given equations.

Revision as of 14:17, 7 May 2020

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.

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