If a, b, c, d, e and f satisfy 2a=3b=6c=9d=12e=18f, then what is the value of \frac{a+b}{c+d+e+f}?

Explanation

Let 2a=3b=6c=9d=12e=18f=k. Then a=\frac{k}{2}, b=\frac{k}{3}, c=\frac{k}{6}, d=\frac{k}{9}, e=\frac{k}{12}, and f=\frac{k}{18}. The numerator is \frac{k}{2} + \frac{k}{3} = \frac{5k}{6} and the denominator is \frac{k}{6} + \frac{k}{9} + \frac{k}{12} + \frac{k}{18} = \frac{15k}{36} = \frac{5k}{12}. Their ratio is \frac{5k/6}{5k/12} = 2.

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