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50 - or more (Posted on 2020-09-10) Difficulty: 4 of 5
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.

No Solution Yet Submitted by Ady TZIDON    
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re(2): some hints & remarks @ Ch | Comment 8 of 14 |
(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 triplet
  
!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
  Posted by Ady TZIDON on 2020-09-11 06:41:50

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