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A friend has given me a puzzle to solve, the puzzle is as follows: $$\color{red}{7 \, , } 24 \to \color{blue}{25}$$ $$\color{red}{12 \, , } 35 \to \color{blue}{37}$$ $$\color{red}{11 \, , } 60 \to \color{blue}{61}$$ $$\color{red}{8 \, , } 15 \to \color{blue}{17}$$ $$\color{red}{9 \, , } 40 \to \color{blue}{\text{M}}$$ then $\color{blue}{\text{M}}$=?

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The answer is

41 because $red^2 + black^2 = blue^2$.

These are all

Examples of Pythagorean triples, and so $9^2 + 40^2 = 81 + 1600 = 1681$, and then $\sqrt{1681} = 41 = M$.

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Although El-Guest answer appears correct for all the listed cases, there could be a simpler answer, which is:

if red is uneven then black + 1 else if red is even then black + 2

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  • $\begingroup$ Huh. I do think that @El-Guest is correct, but this is very interesting. $\endgroup$ – Brandon_J Mar 13 at 13:37
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    $\begingroup$ There are plenty of example cases that would satisfy the intended solution but not this one (e.g., $\color{red}{28 \, , } 45 \to \color{blue}{53}$), and there are plenty of example cases that would satisfy this solution but not the intended one (e.g., $\color{red}{4 \, , } 5 \to \color{blue}{7}$). It's a shame that no case like either of these was in the puzzle to make the solution unambiguous. $\endgroup$ – Joseph Sible Mar 14 at 4:02
  • $\begingroup$ See also this picture (warning: spoiler for the intended solution). The overlap between the two solutions happens to be everything along the left and bottom edges in it. $\endgroup$ – Joseph Sible Mar 14 at 4:14

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