All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
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.

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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
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
4 conc 9=49
sqrt(16) not possible since 16 as concatenation does not exist
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

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information