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The goal of this game is to find the lowest score for the 10 equations below.

0 ? 0 ? 0 ? 0 = 7
1 ? 1 ? 1 ? 1 = 7
2 ? 2 ? 2 ? 2 = 7
3 ? 3 ? 3 ? 3 = 7
4 ? 4 ? 4 ? 4 = 7
5 ? 5 ? 5 ? 5 = 7
6 ? 6 ? 6 ? 6 = 7
7 ? 7 ? 7 ? 7 = 7
8 ? 8 ? 8 ? 8 = 7
9 ? 9 ? 9 ? 9 = 7

The accepted mathematical operators are:
+, −, ×, ÷, x², √, ! ,( and )

Note:
Implicit multiplication isn't allowed (e.g. 2(2))
You are allowed to put one or more of the accepted operators anywhere in the equations, but at least one operator must replace the ?:s.

This is how you calculate the score:
One point for every +, −, ×, ÷, x², √, ! ,( and ), so make the equations as short as possible.
Add together all point for the 10 equations - the lowest score wins!

Test your formula at:
https://www.mathway.com/Algebra

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    $\begingroup$ Unclear.  (1) I guess each ? must be replaced by one or more operator(s)?  Or are you allowing things like (ROT13) bar sbyybjrq ol bar vf ryrira?  (2) I guess we’re allowed to put ( at the beginning and ) at the end.  (3) Is unary minus allowed?  (4) I guess that exponentiation isn’t allowed, although it’s (IMHO) a more fundamental algebraic operator than factorial. Since you have explicitly documented one exclusion (implicit multiplication by concatenation), you should probably mention this. $\endgroup$
    – Peregrine Rook
    Apr 14 '19 at 2:55
  • $\begingroup$ Peregrine Rook, thanks alerting me on that the x² (opposite to √) was missing from the accepted operators, that was not my intention. Sorry, but great job solving it without. I also edited the Note to make it clear that '...one or more of the accepted operators could be anywhere in the equations...' $\endgroup$
    – Ola Ström
    Apr 14 '19 at 9:49
  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful. $\endgroup$
    – Rubio ♦
    Apr 21 '19 at 7:56
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Starting with 10 points for 0...

(0!+0!+0!)!+0!

...and 6 points for 1

(1+1+1)!+1

... and 7 points for 2

(2+2)!!-2/2

... and 3 points for 3

3+3+3/3

... and 3 points for 4

4+4-4/4

... and 5 points for 5

5+(5+5)/5

... and 5 points for 6 (or 6 points if double factorial is not acceptable).

6!!/6-6/6 or 6+√(6*6)/6

... and 5 points for 7

7+(7-7)*7

... and 6 points for 8

8-√(8*8)/8

... and 4 points for 9 (thanks to Jaap Scherphuis)

9-√9+9/9

So total points is 54 points. I haven't found a way to do it without the double factorial for 2.

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  • $\begingroup$ For $6$ and $8$ you can also remove a set of brackets if you extend the square root over the top. $\endgroup$
    – Jaap Scherphuis
    Apr 11 '19 at 6:30
  • $\begingroup$ Penguino, your solution for 8 is great and Jaap, I like your solution for 9. $\endgroup$
    – Ola Ström
    Apr 11 '19 at 12:21
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    $\begingroup$ Ironically, 2 ? 2 ? 2 ? 2 ? 2 = 7 is easier  than 2 ? 2 ? 2 ? 2 = 7. $\endgroup$
    – Peregrine Rook
    Apr 14 '19 at 2:55
  • $\begingroup$ @JaapScherphuis wait, so it wasn't even the radical in the first place that undermined the solution xD Thanks for that pick-up! I can be dumb sometimes... $\endgroup$
    – Mr Pie
    Apr 14 '19 at 9:51
  • $\begingroup$ @Penguino Why didn't you use my suggestion for $6$ and $8$? I.e. use $\sqrt{6*6}$ to eliminate a pair of brackets. $\endgroup$
    – Jaap Scherphuis
    Apr 14 '19 at 11:04
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Perhaps I’m misunderstanding yesterday’s edit to the question, but I believe that I can get the score for 2 down to 4 points:

$2^2+2+2/2$

My understanding of the edit to the question is that, in the N equation, we can use N2 as an operator, and that this counts as only one (not two) of the four uses of N, even if N is 2.

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