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.

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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re(2): Record | Comment 13 of 14 |
(In reply to re: Record by Dej Mar)

Given the limitation to kth factorials where k <= 3 and with the other imposed restrictions of operations, with the assumption that a member of the set of triplets is a single digit number, the limit may be n=210 using the triplets {4, 4, 5}. 

Thinking "outside the box", I realized that the triplets are not stated to be required to be single digits, and with the SQRT operation, the initial set of integers can begin as larger numbers, which can possibly increase the range and limits. For example, instead of {4, 4, 5}, the set was {16, 16, 25}. One could take the SQRT(x) for each member of the triplet and have the same results but with the increased options now available to extend the limit. The process can be done again with squaring of these numbers, eg., {256. 256, 625}, to further increase one's options as one does not need to take the square root for each integer of the set, but may choose to do so for any.  The limit will definitely be increased as one could now reach 211 with
256 - (!(SQRT(SQRT(256))) × (SQRT(SQRT(625))))
= 256 - (9×5) = 211.     
  Posted by Dej Mar on 2020-09-16 14:30:38

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


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information