The abstract and introduction don't explain this very well. My understanding is that the author wrote the digits 1 through 9 in ascending order, and then inserted parenthesis, addition, subtraction, multiplication, division and exponentiation operators between them where appropriate to get every number from 0 to 11111. And then he did the same thing using the digits 9 through 1 in descending order and did the same th…
I wonder if any numbers have multiple solutions.
You can create a second version by replacing this with 1×2×3.
For example,
10914 = 1 × 2 × 3 × (4^5 + 6 + 789)
10914 = (1 + 2 + 3) × (4^5 + 6 + 789)