Fast way to calculate Fraction

Question

I want to solve this problem not by using common denominator. Is there any innovative and fast way ?.

$$ {1 \over 6} + {4 \over 21} + {5 \over 84} + {7 \over 228} + {9 \over 532} =\ {\large ?} $$

Answer 1

You can avoid converting to the single common denominator $1596$ here by grouping terms in a way that creates some nice cancellations:

$$\begin{align} {1\over6}+{4\over21}+{5\over84}+{7\over228}+{9\over532}&={1\over6}+\left({4\over21}+{5\over4\cdot21} \right)+\left({7\over3\cdot76}+{9\over7\cdot76} \right)\\ &={1\over6}+{16+5\over4\cdot21}+{49+27\over21\cdot76}\\ &={1\over6}+{21\over4\cdot21}+{76\over21\cdot76}\\ &={1\over6}+{1\over4}+{1\over21}\\ \end{align}$$

From here it's fairly easy to get

$${1\over6}+{1\over4}+{1\over21}={5\over4\cdot3}+{1\over7\cdot3} ={35+4\over28\cdot3} ={13\over28}$$

Whether this is innovative or fast is hard to say. But I think it beats computing

$${1\cdot266+4\cdot76+5\cdot19+7\cdot3+9\cdot7\over3\cdot4\cdot7\cdot19}$$