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How to use the number 2,0,1,8 to make 51?

You can use any operation except for multiple factorials and you can concatenate numbers, ex.2,0=>20

You can take the square root of a number without a two, but if you want to take the cubic root you will need a three.

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closed as too broad by Ankoganit, Bass, Beastly Gerbil, Rand al'Thor, Glorfindel Jan 9 '18 at 17:49

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ When you say 'multiple factorials' do you mean factorial of a factorial, or just factorials themselves? Also, can the order of the digits be rearranged, or must it be 2 - 0 - 1 - 8 ? $\endgroup$ – Phylyp Jan 9 '18 at 4:14
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    $\begingroup$ You can rearrange them, what I meant was that you can't use double factorials or triple factorials. 8!!=8*6*4*2. $\endgroup$ – Really desperate person Jan 9 '18 at 4:18
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    $\begingroup$ Can we use floor / ceil function? $\endgroup$ – athin Jan 9 '18 at 4:46
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    $\begingroup$ Can we use log? $\endgroup$ – prog_SAHIL Jan 9 '18 at 6:00
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    $\begingroup$ Since no one has said it yet... Welcome on Puzzling.SE! While I personally think there is some potential here, you might notice your questions has been getting some downvotes. This is probably due to the fact that the rules are not very well laid-out - Things like the concatenation, logs and other functions are unclear as to whether they're allowed and in which way. If you edit your question to more complete, you might be getting a better reception. $\endgroup$ – Lolgast Jan 9 '18 at 7:15
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I think this is valid by your rules :

$ \frac{\sqrt{\sqrt{.1^{-8}}}}{2} + 0! $
$= \frac{100}{2} + 1$
$= 51$

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    $\begingroup$ And it doesn't rely on those filthy double factorials! :D (only on a filthy double square root) $\endgroup$ – Keelhaul Jan 9 '18 at 9:54
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$\lceil{(8.1-0!)^2}\rceil=51$

Explanation:

$\lceil$ $\rceil$ rounds above. ; $(8.1-0!)^2=(8.1-1)^2=7.1^2=50.41$ ; $\lceil50.41\rceil=51$

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  • $\begingroup$ I've changed your commas to dots - while commas seem to be the convention in more countries (see wikipedia ), the dot is standard for English speaking countries. Also, the rendering for commas is a bit weird - it inserts spaces, causing the numbers to appear as separate numbers which may leave readers confused at first (including me) $\endgroup$ – Lolgast Jan 9 '18 at 13:35
  • $\begingroup$ @Lolgast You can use {,} to render a decimal comma: $8{,}1$. $\endgroup$ – yo' Jan 9 '18 at 14:24
  • $\begingroup$ Ah, good to know! Though I'm still of the opinion that a dot should be used... That's also the convention my (internationally oriented) university uses for technical, English papers, even though being based in a comma-based country (the Netherlands). $\endgroup$ – Lolgast Jan 9 '18 at 14:26
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Finally got it

Concatenate the following 2 results
(8-2-1)(0!)

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  • $\begingroup$ you can't concatenate numbers after you change them with any o[erations. $\endgroup$ – Really desperate person Jan 9 '18 at 5:11
  • $\begingroup$ That would be to easy. $\endgroup$ – Really desperate person Jan 9 '18 at 5:11
  • $\begingroup$ @Evargalo I think they meant ((8 - 2) - 1), whereas I suspect you mean (8 - (2 + 1)). Same difference :) $\endgroup$ – John Jan 9 '18 at 15:58
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How is this-

(18-0!) * T2 = (18-1) * 3 = 17 * 3 = 51

Here, T is- triangular number

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    $\begingroup$ If we are to use triangular numbers, why not $T(8+0!)+T(2+1)=T9+T3=45+6=51$ ? $\endgroup$ – Evargalo Jan 10 '18 at 8:14
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102/(cube root of 8) = 102/2 = 51

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  • $\begingroup$ You can't take the cube root without a three but square root is allowed without the two. Sorry I should have mentioned that. $\endgroup$ – Really desperate person Jan 9 '18 at 4:45
  • $\begingroup$ In that case, are we allowed to reuse the numbers 2,0,1,8? $\endgroup$ – Mehul Shah Jan 9 '18 at 4:59
  • $\begingroup$ like using 2 more than once? $\endgroup$ – Really desperate person Jan 9 '18 at 5:04
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    $\begingroup$ I didn't make the rules...... $\endgroup$ – Really desperate person Jan 9 '18 at 5:10
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    $\begingroup$ @mehulshah that's because a square root looks like this: $\sqrt{}$, while a cube root looks like this $\sqrt[3]{}$. One has a digit, one doesn't. $\endgroup$ – boboquack Jan 9 '18 at 5:31
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I feel like I'm not too far away:

51 = (2^8-1)/5, but I can't make the /5, a *.2 would also solve it
or, 51 = (8^3-2)*.1, but I can't make the 3. It feels so close, and yet so far...

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$51 = ((2 + 0!)!)!! + \sqrt{(1 + 8)}$

$= 6!! + 3 = 2 * 4 * 6 + 3 = 48 + 3$

!! is Double factorial.

Credit also goes to @KaiNoack for his comment on my question and to EmNero from the German math forum who used double factorials to get 48.

EDIT: I've just noticed I misread the comment by @ReallyDesperatePerson. I thought it said you can use double factorials, but it says you can't use them. Sorry! I hope I can leave the answer because it is mathematically correct, but it is against the challenge rules so please don't upvote it.

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  • $\begingroup$ Multiple factorial is prohibited. $\endgroup$ – athin Jan 9 '18 at 10:00
  • $\begingroup$ Oh. I've just noticed I misread the comment by @ReallyDesperatePerson. I thought it said you can use double factorials, but it says you can't use them. Sorry! $\endgroup$ – fitch496 Jan 9 '18 at 10:04
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solution : √(2801-200) = √2601 = 51

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  • $\begingroup$ you can't use a number more than once $\endgroup$ – Really desperate person Jan 9 '18 at 5:31
  • $\begingroup$ @Reallydesperateperson Then why did you accept an answer that used the digits tons and tons of times? $\endgroup$ – JLRishe Jan 24 '18 at 17:55

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