We know that the sum of all the portions have to be equal to 1. Let x be the missing part, then we have that:
[tex]\begin{gathered} x+\frac{1}{12}+\frac{1}{9}+\frac{1}{6}+\frac{1}{4}+\frac{1}{3}=1 \\ x+\frac{9+12}{108}+\frac{4+6}{24}+\frac{1}{3}=1 \\ x+\frac{21}{108}+\frac{10}{24}+\frac{1}{3}=1 \\ x+\frac{7}{36}+\frac{5}{12}+\frac{1}{3}=1 \\ x+\frac{7}{36}+\frac{15+12}{36}=1 \\ x+\frac{7}{36}+\frac{27}{36}=1 \\ x+\frac{34}{36}=1 \\ x+\frac{17}{18}=1 \\ x=1-\frac{17}{18} \\ x=\frac{1}{18} \end{gathered}[/tex]
Therefore the missing portion is 1/18.
