# prove that 2016 is a self-composable number

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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Can I do things like (20)^(16) ? – Diego Martinoia Feb 22 at 10:44
no, concatenation of digits is not allowed since otherwise you might express 2016 as.... 2016 :-) – mau Feb 22 at 13:26
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. – mau Feb 22 at 13:35
What are the extended operations? – marmistrz Feb 22 at 21:31

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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Seems good to me. Good job. And here I was trying to write a graph explorer... – Diego Martinoia Feb 22 at 12:24
you are right! (Ok, it was not easy) – mau Feb 22 at 13:26
Ah, relies on the (somewhat arbitrary) rule that 0! = 1. I've always felt that it should simply be undefined for numbers < 1... – Darrel Hoffman Feb 22 at 15:07
@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). – kasperd Feb 22 at 15:18
I don't know if $.\overline{0!}$ is actually valid mathematical notation... – GentlePurpleRain Feb 22 at 17:18

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