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Define a number self-composable if it may be computed using just the digits of the number itself (used just once) and the following operations:

  • basic operations ( $+, -, \times, \div$)
  • less than basic operations ($x^y, \;\sqrt x, \;x!, \;x.y$ (decimal point))
  • extended operations ($.x, \; .\overline{x}$)
  • parentheses at will

If the digits of the mathematical operation are in the same order as in the number itself, the number is said orderly self-composable.

For example, 25 is self-composable ($5^2 = 25$) and 343 is orderly self-composable, since $(3+4)^3 = 343$.

2016 is self-composable too: find how.

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  • $\begingroup$ Can I do things like (20)^(16) ? $\endgroup$ – Diego Martinoia Feb 22 '16 at 10:44
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    $\begingroup$ no, concatenation of digits is not allowed since otherwise you might express 2016 as.... 2016 :-) $\endgroup$ – mau Feb 22 '16 at 13:26
  • $\begingroup$ Let's say that concatenation is allowed except for the trivial case in which you are writing the number that you start with: anyway in this case it is not necessary. $\endgroup$ – mau Feb 22 '16 at 13:35
  • $\begingroup$ What are the extended operations? $\endgroup$ – marmistrz Feb 22 '16 at 21:31
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I think I have it

$ 2016 = ({.2} / {.\overline1} + 0!) \times 6!$

Also as pointed out by Matt in the comments, we can swap things around so that 2016 is orderly self-composable

$ 2016 = ({.2} / .\overline{0!} + 1) \times 6!$

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    $\begingroup$ Seems good to me. Good job. And here I was trying to write a graph explorer... $\endgroup$ – Diego Martinoia Feb 22 '16 at 12:24
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    $\begingroup$ @DarrelHoffman That is not arbitrary at all. By definition we have $n! = (n - 1)! * n$. And if we take $n=1$ we get $1! = 0!$. There is even a (much more involved) generalization to all real numbers (except negative integers). $\endgroup$ – kasperd Feb 22 '16 at 15:18
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    $\begingroup$ I don't know if $.\overline{0!}$ is actually valid mathematical notation... $\endgroup$ – GentlePurpleRain Feb 22 '16 at 17:18
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    $\begingroup$ @DarrelHoffman You may be interested in math.stackexchange.com/questions/20969/… . The many answers there, including mine at math.stackexchange.com/a/1094926/198422 , suggest a lot of reasons why you'd want $0!$ to equal $1$. I would say that the equation $0! = 1$ is conventionally accepted among the mathematics community: just see the link; there's not really any dissent as to what the "right value" of $0!$ should be. $\endgroup$ – mathmandan Feb 22 '16 at 18:36
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    $\begingroup$ $0!=1$, but $.\overline{0!}$ is really strange. $\endgroup$ – f'' Feb 22 '16 at 19:09
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find divisors of 2016 that contains its own numbers: $2016/2 \rightarrow 1008/2 \rightarrow 504/2 \rightarrow 252/2 \rightarrow 126/126 \rightarrow 1$

therefore;

we can't use $2^4$ since $4$ is not in $2016$ but we can do $2^2$ for $2$: $2^2 \times 2^2 \times 126 \times 1^0$

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  • $\begingroup$ in your solution you are using the 2 more than once, and you are concatenating digits. Both of these are not allowed. $\endgroup$ – mau Feb 22 '16 at 9:53
  • $\begingroup$ oh I see. missed the point "just once" will think some more :) $\endgroup$ – canova Feb 22 '16 at 9:55

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