The solution of such problem is a complete list of the combinations, like in "four fours, “five fives", resolving 2023, 1234 et al.
Let as call the first unachievable number a Nova number or a Nn. Thus Nn(four fours) at present time is 113, until some solver "makes it better" (Hey, Jude!). From now on in the problems relating to generation of the final list, the solver should only post his Nn until another solver improves it. Now i present as an example few exercises and request the following procedure: 1st solver say A publishes his Nn and if nobody improves it within 2 weeks then A’s list and his Nn will be considered an official solution. If another person, say B claims a higher Nn then the above procedure is followed.
to 1st assignment: Four fours.
113 = (44/4) & sqrt(!4) using concatenation (&) for the first time and the !4 to denote subfactorial, thus adding additional list of achievable numbers.
Since
120 =(4+4/4)!
121=(44/4) ^ sqrt4
122= 120+sqrt4 etc
Evaluate the Nn{4,4,4,4}
allowing use of concatenation, sqrt, subfactorials etc ( no operator used more than twice.
Second assignment
What generating quadruple has higher Nn : {2,3,5,8} or {2,5,7,9)? Same constraints as in 4 fours.
Third challenge ( optional): Create and post your own puzzle based on Nn, not necessarily assuming the same rules,
very good
