An old numerical problem that keeps reappearing in puzzle books as though it had never been analyzed before is the problem of inserting mathematical signs wherever one likes between the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 to make the expression equal 100. The digits must remain in ascending order. There are many hundreds of solutions, the easiest to find perhaps being:
1 + 2 + 3 + 4 + 5 + 6 + 7 + (8 × 9) = 100
The problem becomes more of a challenge if the mathematical signs are limited to plus and minus. Here again there are many solutions, for example:
1 + 2 + 34 − 5 + 67 − 8 + 9 = 100
12 + 3 − 4 + 5 + 67 + 8 + 9 = 100
123 – 4 – 5 – 6 – 7 + 8 – 9 = 100
123 + 4 – 5 + 67 – 89 = 100
123 + 45 – 67 + 8 – 9 = 100
123 – 45 – 67 + 89 = 100
“The last solution is singularly simple,” writes the English puzzlist Henry Ernest Dudeney in the answer to puzzle number 94 in his Amusements in Mathematics, “and I do not think it will ever be beaten.” As far as I know, Dudeney’s claim has never been challenged.
In view of the popularity of this problem, it is surprising that so little effort seems to have been spent on the problem in reverse form. That is, take the digits in descending order, 9 through 1, and form an expression equal to 100 by inserting the smallest possible number of plus or minus signs.
The answer was still open as of the October 1967 issue of Scientific American. But Martin Gardner promised to reveal the best solution he knew in the subsequent issue. How few signs can you solve it with?
To form an expression equal to 100, plus and minus signs can be inserted between the digits, taken in reverse order, as follows: 98 – 76 + 54 + 3 + 21 = 100. Did any reader manage to do it with fewer signs?
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A version of this puzzle originally appeared in the October 1962 issue of Scientific American.