The number of solutions $(x,y,z)$ to the system of equations

$x+2y+4z=9$

$4yz+2xz+xy=13$

$xyz=3$

such that at least two of $x,y,z$ are integers is

The answer is 5.

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Note that $(T-x)(T-2y)(T-4z) = T^3-9T^2+26T-24 = (T-2)(T-3)(T-4).$

So $(x,2y,4z)$ is a permutation of $(2,3,4)$ . Five of these six permutations lead to solutions where at least two are integers (the exception is $(4,3,2)$ ). So the answer is $\fbox{5}$ .