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There are many number sequence puzzles on this site. I acknowledge them all.

Martin Gardner introduced the puzzle with the 9 digit sequence with math operations, reaching 100 as below

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, 9 to make the expression equal 100. The digits must remain in the same sequence. There are many hundreds of solutions, the easiest to find perhaps being "

1 + 2 + 3 + 4 + 5 + 6 + 7 + ( 8 x 9 ) = 100

With concatenation allowed the fewest possible operations to get to 100 is

123-45-67+89=100 which uses only 3 operations

Based on that

Using 1 2 3 4 5 6 7 8 9 and fewest math operations (4 or less), with parentheses counting as operations, can you get 0?

AND do the same for Reverse order

9 8 7 6 5 4 3 2 1 = 0

Your answers must represent all the digits in sequence. All digits must appear once. Only + - x / ! allowed. Parentheses are counted as though they were operations; each pair of parentheses is 2 operations. Every sign introduced is a step. No partial answers. I expect more than one solutions. I have 4 step solutions but there may be 3 step solutions too.

Hope the moderators will not consider this as an open question

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  • $\begingroup$ can we use a computer to solve this puzzle? $\endgroup$
    – melfnt
    Commented Feb 12, 2020 at 16:58
  • $\begingroup$ You can but I found it much more fun to do this on paper. There are logical deductions. $\endgroup$
    – DrD
    Commented Feb 12, 2020 at 17:05
  • $\begingroup$ Is $(1)$ consider as two operations (two parenthesis) or is it only one since you need a right parenthesis for every left one? If it is the later, there is an easy 4 steps solution. $\endgroup$
    – Alain Remillard
    Commented Feb 12, 2020 at 17:09
  • 1
    $\begingroup$ Does (x+y)z count as 2 or 3? Given the * is implied in standard notation $\endgroup$
    – Gamora
    Commented Feb 12, 2020 at 17:29
  • 4
    $\begingroup$ $1^{234567}+8-9=0$ - I'm not sure if this is allowed. $\endgroup$
    – ThomasL
    Commented Feb 12, 2020 at 22:08

5 Answers 5

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With 4 symbols.

For the normal sequence:

1*23-45-67+89

For the reverse sequence:

98-76-54+32*1

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This should be all solutions with 4 operations. For the forward sequence:

$$ 1*23 - 45 - 67 + 89 = 0 $$

For the backward sequence:

$$ 987 - 6! + 54 - 321=0\\98*7 - 654 - 32/1 = 0\\ 98*7 - 654 - 32*1 = 0\\ 98 - 76 - 54 + 32/1 = 0\\ 98 - 76 - 54 + 32*1 = 0$$

The first three variations are not in any of the other answers :)

There is another variant for the forward sequence with factorial (5 steps)

$$1 + 23 + 45 + 6! - 789 = 0$$

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  • $\begingroup$ Good job on the backward sequence @daw. Is there a 3 step solution? $\endgroup$
    – DrD
    Commented Feb 13, 2020 at 15:32
  • $\begingroup$ @DEEM Found another with factorial. But no 3 step solution $\endgroup$
    – daw
    Commented Feb 13, 2020 at 15:43
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A way to do it with $\color{red}{4}$ steps, considering parenthesis comes in pair (one step for the pair)

OP clarified that each individual parenthesis count so $5$ steps.

$$(1+2-3)*456789=0$$ $$987654*(3-2-1)=0$$

Not part of the solution, since the floor function is not allowed, but it could be done in $3$ step. Where $\lfloor x\rfloor$ is the greatest integer smaller than $x$, e.g. $\lfloor 0.1\rfloor=0$.

$$\lfloor1\div23456789\rfloor=0$$ $$\lfloor9\div87654321\rfloor=0$$

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5 operations

$(1+2-3) * 456789 = 0$
$987654 * (3-2-1) = 0$

4 operations

$98-76-54+32*1$

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Using 5 operations:

$(12/3-4)*56789=0$

Using 4 operations:

$9 8- 7 6 -5 4+3 2/1 = 0$

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