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

Revision as of 14:45, 7 May 2020

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

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

@@ Line 14: / Line 14: @@
 == 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 <cmath>(x+*)(y+*)(z+*)</cmath> where the blanks should be filled in with numbers.
-Filling in as <cmath>(x+4)(y+2)(z+1)</cmath> gives <cmath>2x+4y+8z+xy+4yz+2xz+xyz+8</cmath>, and all parts happen to be multiples of the given equations. After substitution, we get <cmath>(x+4)(y+2)(z+1)=60</cmath>.