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 50 - or more (Posted on 2020-09-10)
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.)
 re: another, higher............. and higher | Comment 4 of 14 |
(In reply to another, higher by Charlie)

I WAS BLIND BUT NOW I SEE

92= 94-2
93= 94-1
94= 94
95= 94+1
96= 94+2
97= 94+3
98= 49*2
99=99
100=99 +1
101=99 +2
102=Not possible Np
103=99 +4
104=Np
105=Np
106= Np
107=99+4!!

TBC BY Ch

btw 71=99-18= 99-((sqrt(!4))!)!!!

Edited on September 12, 2020, 7:05 am
 Posted by Ady TZIDON on 2020-09-10 17:45:12

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