Representing Numbers With Digits

My previous post focused on the number 1089 in which I made reference to a newly discovered blogging site at https://math1089.in/ and in particular to a post about the number 1089. In that post it was noted that:$$\begin{array}{rl}1089& =12\times 3^4+5\times 6+78+9\\ 1089& =987+65+4+32+1\end{array}$$In another post about the number 108, it was noted that:$$\begin{array}{rl}108& =1+2+3+4+5+6+78+9\\ 108& =9+8\times 7+6\times 5+4\times 3+2‒1\end{array}$$This got me thinking about whether it was possible to represent every number at least once in terms of consecutive single or concatenated digits separated by the basic operations of arithmetic combined with exponentiation and brackets.

For example, my diurnal age today is 27339. Is such a feat possible for this number? The determination is not easy because there are just  so many possible ways to combine the digits from 1 to 9. I spent quite some time playing around with the possibilities and I did come close but not close enough. The exercise is oddly addictive. There's no serious mathematics involved in the exploration. It's more in the nature of a puzzle, like Sudoku. 

There are a variety of strategies, one of which is to establish base points with as few digits as possible. An example of this is:$$35016=1+2+3+4+5+6^7/8+9$$Here the digits 6, 7 and 8 combine to form 34992 and the remaining digits are free to be manipulated. It's true that we can use 34567 to get to a similar number but here five digits are tied up and there's little that can be done with the remaining digits (1, 2 and 8, 9) since they are separated. Here is an example of how we can get close to 27339 using the earlier mentioned base point:$$28084=-({12}^3\times 4+5)+6^7/8+9$$Another approach is to use factors. We know that 27339 = 3 x 13 x 701 and it's easy enough to create the first two factors using 1 + 2 = 3 and 3 x 4 - 5 + 6 = 13. However, we are then left with 789:$$\begin{array}{rl}27339& =3\times 13\times 701\\ 30771& =(1+2)\times (3\times 4-5+6)\times 789\\ & =3\times 13\times 789\end{array}$$While this works fine for 30771, it's not of much use for 27339. In general, this approach is quite restrictive and the more promising approaches will involve additions and subtractions along with exponentiation, multiplication and division.

At the moment I don't have a solution to the specific problem of representing 27339 in terms of sequential digits and I certainly don't have an answer to the general problem of whether such a representation is always possible or only sometimes possible. Certainly there's an upper limit on the number size and this is imposed by the nine digit restriction but such a limit is huge and my investigation is focusing on numbers in the region of 30000. My suspicion is that it's not always possible within the restrictions imposed. If we relax the requirement that the digits need to be in sequential order or we allows square roots, factorials etc. then maybe it's possible but for the moment I'll keep within the earlier rules and keep revisiting the problem from time to time.