Sharing this great and fun math and critical thinking puzzle with students is the best way to get them thinking mathematically.

## Challenge:

There are 4 fractions in the figure above with the same difference. What is the sum of all fractions?

**A-**\(\frac{1}{126}\)

**B-** \(\frac{7}{33}\)

**C-** \(\frac{13}{42}\)

**D-** \(\frac{10}{66}\)

**E-** \(\frac{13}{21}\)

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The correct answer is E.

There are four fractions with the same difference. First, we need to find the missing fractions.

The difference of \(\frac{1}{7}\) and \(\frac{1}{6}\) is:

\(\frac{1}{6} – \frac{1}{7}/7 = \frac{7-6}{42} = \frac{1}{42}\)

The difference of \(\frac{1}{7}\) and the first fraction is:

\(\frac{1}{42} ÷3 = \frac{1}{126}\)

The first missing fraction is:

\(\frac{1}{7} + \frac{1}{126}= \frac{18+1}{126} = \frac{19}{126}\)

The second missing fraction is:

\(\frac{19}{126} + \frac{1}{126}= \frac{20}{126}\)

The sum of all fractions equals:

\(\frac{1}{7} + \frac{19}{126}+ \frac{20}{126} + \frac{1}{6} = \frac{18}{126}+ \frac{19}{126} + \frac{20}{126} + \frac{21}{126} = \frac{78}{126} = \frac{13}{21}\)