In the spirit of the classic four fours, I wonder what's the optimal set of four numbers?
Your goal is to make the most consecutive integers using four digits of your choice. Pick four: $0,1,2,3,4,5,6,7,8,9$ ( You can pick multiple instances of the same digit )
When constructing an integer:
- all of your four candidates must be used exactly once (order/placing of digits is irrelevant)
- You may use basic arithmetic operations $+,-,\times,\div$ and parentheses $()$
- You may use $a^b$ and $\sqrt[a]{b}$ but at the expense of 2 numbers as you can see
- You may not form new numbers, i.e. $ab$ is not allowed
If we were to use four $4$s, the best we could do would be up to $9$:
0 = 4 ÷ 4 × 4 − 4
1 = 4 ÷ 4 + 4 − 4
2 = 4 −(4 + 4)÷ 4
3 = (4 × 4 − 4)÷ 4
4 = 4 + 4 ×(4 − 4)
5 = (4 × 4 + 4)÷ 4
6 = (4 + 4)÷ 4 + 4
7 = 4 + 4 − 4 ÷ 4
8 = 4 ÷ 4 × 4 + 4
9 = 4 ÷ 4 + 4 + 4
*10 = 4 ÷√4 + 4 ×√4
*10 = (44 − 4) ÷ 4
Number 10 can't be done and is an example of failing, since it would require either:
expenseless roots; $\sqrt{4}$ isn't allowed. ( $\sqrt[2]{4}$ is, which requires you to use $4$ and $2$ )
number formation which isn't allowed either.
Zero does not necessarily need to be included, you can start at either $0$ or $1$.
For the purposes of freedom of puzzling, if you think you can top your solution for a chosen set of digits, by starting at any other positive integer, you can add that to your answer below your initial solution. (I suspect this is unlikely)
If you want, you can extend your consecutive list to negative integers but this is strictly optional and not necessary in any way, other than for the purposes of fulfillment and mathematical euphoria.
Example
There is an example on Puzzling.SE using digits $2,2,4,5$:
But this can be expanded since the given example uses only basic arithmetic operations, not including potentiation and roots. I also suspect It could be done better using another set.
I tried this by hand and I'm stuck at number $29$ using this example set, and at number $34$ using $9,8,3,2$.