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Previous huge out of tiny puzzle by TerriblyDrawn

Using only four ones, four twos, and five fives (1, 1, 1, 1, 2, 2, 2, 2, 5, 5, 5, 5, 5), can you make the complex number$$51+5i\sqrt5$$You can use all mathematical operations you know (i.e floor, factorial, multifactorial, powers/indicessee restriction 1, etc.), but concatenation is not allowed (e.g. $11$ from $1$ and $1$).

Some restrictions:

  1. Any powers used must be derived from the initial set (i.e $\sqrt5\implies5^{1/2},4^3\implies(2\cdot2)^{1+2}\text{ or }(5-1)^{5-2}$)
  2. You must use all of the numbers (I honestly think it's only possible this way)

Notes

  1. Sorry for deriving away from the set of numbers used in the other two puzzles, I just wanted to be a bit creative, but I definitely understand if you are unhappy with this.
  2. Here are some more examples for restriction 1 if that is okay:$$(-3)^3\implies(2-5)^{5-2},((1-2)(1+2))^{5-2}\\\sum_{9\ge k\ge 0}k^{12}\implies\sum_{(2+1)^2\ge k\ge1-1}k^{2\,\cdot\,5+2}\\(251!!)^{34}\implies((5\cdot5\cdot5\cdot2+1)!!)^{5\,\cdot\,5+1+1+1+2+2+2}$$
  3. Final note: I have checked, and there honestly might only be one solution tbh.
  4. Sorry for not saying this beforehand, but $.1,.2,.5$ is allowed with only one number, but $1.1,1.2,1.3$ wouldn't be due to concatenation.
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    $\begingroup$ Interesting restriction on √ not allowed. $\endgroup$
    – Weather Vane
    Oct 21 at 19:30
  • $\begingroup$ @WeatherVane It's because$$a^{1/2}=\sqrt a\because a^{1/b}\gets\sqrt[b]a\quad\forall b\in\mathbb R,b\ne0$$so I just don't want anyone abusing that power. $\endgroup$
    – CrSb0001
    Oct 21 at 19:32
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    $\begingroup$ Also, are we allowed to use $i$? Or must it be $-1^{.5}$ ? $\endgroup$
    – Weather Vane
    Oct 21 at 19:36
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – CrSb0001
    Oct 21 at 19:37

2 Answers 2

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Here is my solution:

$(5 \times 5 \times 2 + 1) + 5 \times (-1)^{1 / \lfloor\frac{5}{2}\rfloor} \times 5^{\frac{1}{2}} = 51 + 5 \sqrt{-1} \sqrt{5} = 51 + 5i \sqrt{5}$

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  • $\begingroup$ ...aaaaand the green checkmark is yours! $\endgroup$
    – CrSb0001
    Oct 21 at 19:55
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    $\begingroup$ Cheers! Also, there's definitely more than one solution since $((2+1)!)!! + 2 + 1 + 5 \times 5^{1/2} \times (\frac{-5}{5})^{2/(5-1)}$ also does the trick! $\endgroup$
    – Bryce
    Oct 21 at 20:02
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I think this should do the trick.

$(5*5*2+1) + (1-2)^{(1-(5/2))} * 5^{(5/2-1)}$

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  • $\begingroup$ While that does simplify to the correct answer, not only did Bryce post faster originally, Bryce also edited their post to show that they got the correct answer more quickly. But otherwise, good job! $\endgroup$
    – CrSb0001
    Oct 21 at 19:55
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    $\begingroup$ @CrSb0001 I noticed Bryce posted while I was formatting my answer, so he definitely deserves the tickmark. However since the answer is different from his I will leave it anyway $\endgroup$
    – Bart-Jan van Rossum
    Oct 21 at 19:57

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