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This puzzle has been bugging me for quite a few days now. I came across this puzzle in an online website.

2, 4, 1 = 4
3, 1, 6 = 8
7, 2, 4 = 7
1, 0, 8 = ?

Can someone please give a hint. Thanks!

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  • $\begingroup$ Why the downvotes? $\endgroup$
    – Charlie
    Commented Jul 21, 2022 at 10:00
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    $\begingroup$ Downvotes are probably due to attribution issues. This puzzle may be on topic, but since you clearly stated that you found it online, we require a bit more detail about the source. $\endgroup$ Commented Jul 21, 2022 at 14:16
  • $\begingroup$ There's a banner along the top of your question with information about how to fix your question. $\endgroup$
    – bobble
    Commented Jul 21, 2022 at 22:39
  • $\begingroup$ oh, ok. it is from an app called Math Riddles $\endgroup$
    – Charlie
    Commented Jul 22, 2022 at 13:14

1 Answer 1

3
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Hint

The pattern involves 4 operations: addition, concatenation, swapping and one more.

Answer

The answer is

$9$

because

$\DeclareMathOperator{\swap}{swap}\DeclareMathOperator{\concat}{concat}$ For numbers $a, b, c = d$, we notice that $$\sqrt{a + \concat(\swap(b, c))} = \sqrt{a + \concat(c,b)} = \sqrt{a + 10c + b} = d$$
For the given case, $$\sqrt{1 + \concat(\swap(0, 8))} = \sqrt{1 + \concat(8, 0)} = \sqrt{1 + 80} = \sqrt{81} = 9$$ This pattern is consistent with the listed numbers: \begin{align} \sqrt{2 + \concat(\swap(4, 1))} = \sqrt{2 + \concat(1, 4)} = \sqrt{2 + 14} &= \sqrt{16} = 4\\ \sqrt{3 + \concat(\swap(1, 6))} = \sqrt{3 + \concat(6, 1)} = \sqrt{3 + 61} &= \sqrt{64} = 8\\ \sqrt{7 + \concat(\swap(2, 4))} = \sqrt{7 + \concat(4, 2)} = \sqrt{7 + 42} &= \sqrt{49} = 7\\ \end{align}

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  • $\begingroup$ Please do not answer questions which are off-topic per our attribution policy; that policy is in place to prevent cheating and plagiarism. $\endgroup$
    – bobble
    Commented Jul 21, 2022 at 22:38
  • $\begingroup$ @bobble As a new user on this site, I wasn’t aware. Thanks for the heads up, will keep that in mind for the future. $\endgroup$
    – user80184
    Commented Jul 21, 2022 at 22:39
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    $\begingroup$ No problem! Feel free to drop into chat for a more informal environment - the regulars there can talk you through how the site works if you have more questions about rules/history. $\endgroup$
    – bobble
    Commented Jul 21, 2022 at 22:41

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