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Found this question in a competitive math test for elementary students. The long way is to add all the decimal values but is there a pattern/trick to solve this question (or these types)? I don't know how to solve this except by the long method of adding all the decimal equivalents.

The Answer is $1$

Compute: $$\frac17 + \frac18 + \frac19 + \frac1{10} + \frac1{11} + \frac1{12} + \frac1{14} + \frac1{15} + \frac1{18} + \frac1{22} + \frac1{24} + \frac1{28} + \frac1{33} = ? $$

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  • $\begingroup$ Well, you could get them to common denominators - I don't know if that would be quicker, though. $\endgroup$ – heather Sep 15 '16 at 17:27
  • $\begingroup$ the searched sum is $1$ $\endgroup$ – Dr. Sonnhard Graubner Sep 15 '16 at 17:29
  • $\begingroup$ The smallest common denominator is $2^3\cdot5\cdot3^2\cdot7\cdot11$. $\endgroup$ – barak manos Sep 15 '16 at 17:30
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Not sure if this is what you are looking for but this is what I did:

$$S = \frac17 + \frac18 + \frac19 + \frac1{10} + \frac1{11} + \frac1{12} + \frac1{14} + \frac1{15} + \frac1{18} + \frac1{22} + \frac1{24} + \frac1{28} + \frac1{33}$$

I observed the following groupings:

  1. 7, 14, 28
  2. 8, 12, 24
  3. 11, 22, 33
  4. I did grunt work on the remaining four numbers and found that the LCM is $90$

$$S = \underbrace{\frac17 + \frac1{14} + \frac1{28}}_{\frac14} + \underbrace{\frac18 + \frac1{12} + \frac1{24}}_{\frac14} + \underbrace{\frac1{11} + \frac1{22} + \frac1{33}}_{\frac16} + \underbrace{\frac19 + \frac1{10} + \frac1{15} + \frac1{18}}_{\frac13}$$ $$S = \frac14 + \frac14 + \frac16 + \frac13$$ $$ = \frac12 + \frac12 = 1$$

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