# Solution

(related to Problem: The Digital Century)

There is a very large number of different ways in which arithmetical signs may be placed between the nine digits, arranged in numerical order, so as to give an expression equal to $100.$ In fact, unless the reader investigated the matter very closely, he might not suspect that so many ways are possible. It was for this reason that I added the condition that not only must the fewest possible signs be used, but also the fewest possible strokes. In this way, we limit the problem to a single solution, and arrive at the simplest and therefore (in this case) the best result.

Just as in the case of magic squares there are methods by which we may write down with the greatest ease a large number of solutions, but not all the solutions, so there are several ways in which we may quickly arrive at dozens of arrangements of the "Digital Century," without finding all the possible arrangements. There is, in fact, the very little principle in the thing, and there is no certain way of demonstrating that we have got the best possible solution. All I can say is that the arrangement I shall give as the best is the best I have up to the present succeeded in discovering. I will give the reader a few interesting specimens, the first being the solution usually published, and the last the best solution that I know.

$$\begin{array}{rrr} &\text{Signs}&\text{Strokes}\\ 1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 \times 9) = 100&9&18\\ (1 \times 2) - 3 - 4 - 5 + (6 \times 7) + (8 \times 9) = 100&12&20\\ 1 + (2 \times 3) + (4 \times 5) - 6 + 7 + (8 \times 9) = 100&11&21\\ (1 + 2 - 3 - 4)(5 - 6 - 7 - 8 - 9) = 100&9&12\\ 1 + (2 \times 3) + 4 + 5 + 67 + 8 + 9 = 100&8&16\\ (1 \times 2) + 34 + 56 + 7 - 8 + 9 = 100&7&13\\ 12 + 3 - 4 + 5 + 67 + 8 + 9 = 100&6&11\\ 123 - 4 - 5 - 6 - 7 + 8 - 9 = 100&6&7\\ 123 + 4 - 5 + 67 - 8 - 9 = 100&4&6\\ 123 + 45 - 67 + 8 - 9 = 100&4&6\\ 123 - 45 - 67 + 89 = 100&3&4 \end{array}$$

It will be noticed that in the above I have counted the bracket as one sign and two strokes. The last solution is singularly simple, and I do not think it will ever be beaten.

Thank you to the contributors under CC BY-SA 4.0!

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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