Let's start with a triplet of integers, say (1, 2, 5) and a set of mathematical operations (+, -, *, /, ^, sqrt, fact!, concatenation, brackets).

Our task will be to represent all (or *almost all* - as explained below) integers from 1 to n using some or all of the initial triplet and any quantity of operations defined above.

So:

1=1

6=1+5

9=5*2-1

13=15-2

27=51-4!

60=12*5 etc

Let's define n as the first occurrence of not being able to find a valid representation for n+1 and for n+2. I believe that in our case n=17 (15+2), since neither 18 nor 19 get valid solutions.

**
You are requested to find a triplet of integers (a,b,c) enabling a maximal n.**

(In reply to

re: some hints & remarks @ Ch by Dej Mar)

tnx 4 ur q's

I wanted to enter some clarifications,* inter alia, *answering most of your questions so the rules would be better defined. Alas, the puzzle was pushed and published - so the best I can do is to address all your issues and ask the solvers to fill in the existong"holes"

in order to improve the cuurrent solution i.e. **105=99+6.**

**Please consider it as a final amendment adding the missing clarifications to the original text.**

**Now , my answers.**

! is a mathematical operator and may be used to imply any of the family of factorial operations that use the factorial symbol operator (!) -- ** including** all of the factorial family members using the ! sign ( subfactorials and double or even triple factorials)** excluding ** primorials and superfactorials.** fact!** was misleading

The radical sign means sqrt of the expression under the sign .

It may** not ** be concatenated with a superscripted integer n to form an nth root.

The concatenation applies only to positive integers, not expressions.

**Examples: ....assuming 4,9,9 was the chosen triple**t

!3=2

!2=1

4 conc 9=49

sqrt9=3

sqrt(49)=7

sqrt(16) not possible since 16 as concatenation does not exist

sqrt(1+6+9)=sqrt((!sqrt4+(sqrt9)!+9))=4

sqrt4 conc sqrt9 is meaningless, neither 49 nor 23

usually one can write 94, 99 , 49 but neither 29 nor 61

For all other operations (concatenation excluded) you are allowed to pick up 3 or less numbers from the following list:

1,1,1,2,2,2,3,3,4,6,6,'8,9,9, since

3,6,9 are derived from 9

4 ,8 from 4

1 & 2 from either 9 or 4

5 , 7 cannot be created from a single digit , be it 4 oe 9

18=9+9 BUT NOT 6+6+6 -2 sixes are generated by nines , the 3rd cannot be created from 4.

Any otherv questions? Do not hesitate to inquire, I will respond.

Rem: I'm not entirely sure that 4,9,9 is rhe best choiice, maybe 5,9,9 or 7,9,9 or 5,5 is preferable. ** ????Who knows??**

*Edited on ***September 11, 2020, 8:29 am**