# 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.

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

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

• @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). Feb 22 '16 at 15:18
• I don't know if $.\overline{0!}$ is actually valid mathematical notation... Feb 22 '16 at 17:18
• @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. Feb 22 '16 at 18:36
• $0!=1$, but $.\overline{0!}$ is really strange.
– f''
Feb 22 '16 at 19:09
• @DarrelHoffman $0!$ would be the product of everything in the empty set. The only meaningful definition of the product of everything in the empty set is $1$ just like the sum of everything in the empty set has to be $0$. The recursive definition of $n!$ does need a basis somewhere, and that basis is defining $0!$ because that's the smallest integer for which it makes sense. It cannot be extended to $-1!$ because that would require a division by zero. It can however be extended to have a meaning for all reals except the negative integers. Feb 22 '16 at 19:30

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 $2$: $2^2 \times 2^2 \times 126 \times 1^0$

• 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 '16 at 9:53
• oh I see. missed the point "just once" will think some more :) Feb 22 '16 at 9:55