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Can you use each number 1, 2, 3, 4, 5, 6, 7 exactly once, the four operations +, -, *, / and the parentheses to construct the number 2024? Bonus: can you find multiple distinct solutions? No computers please.

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  • $\begingroup$ The last 3 questions of that format where all downvoted and closed. Unless there is some clever twist on the concept, the idea of 'make this number from these digits with some basic operations' is just not that interesting. $\endgroup$
    – quarague
    Commented Jan 14 at 15:30
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    $\begingroup$ I thought the same when I saw the title. But unlike the other questions that requested a whole range of values which were way too easy to produce, given too many operators, this question requests a single result and the solution is far from obvious to find. It is not the same level. $\endgroup$
    – Florian F
    Commented Jan 14 at 18:16
  • $\begingroup$ What is considered 'distinct' could use some more explanation. (e.g. is the swap Florian F suggests a distinct solution?; is 3x2 distinct from 2x3?) Now it is unclear how many distinct solutions there are. $\endgroup$
    – Retudin
    Commented Jan 14 at 19:08
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    $\begingroup$ Commutativity should not count as distinct imho. But I had another swap in mind. $\endgroup$
    – Florian F
    Commented Jan 14 at 21:40

1 Answer 1

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One solution would be

$2024 = (2 + 3 \times 4 \times 6 \times 7) \times (5 - 1)$

A significantly different one is

$2024 = (1 + (2 + 4) \times 6 \times 7) \times (3 + 5)$

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  • $\begingroup$ There is an easy swap for a second solution. $\endgroup$
    – Florian F
    Commented Jan 14 at 18:06
  • $\begingroup$ @FlorianF I'm aware. Just not sure it should count as different. $\endgroup$ Commented Jan 14 at 19:57
  • $\begingroup$ Well done. That's the first solution. Can you find one more? The direct swap does not count. $\endgroup$ Commented Jan 15 at 4:00
  • $\begingroup$ @DmitryKamenetsky I think I found it. It's related but not trivially so. $\endgroup$ Commented Jan 15 at 4:57
  • $\begingroup$ Yes that's the one, congratulations! $\endgroup$ Commented Jan 15 at 6:57

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