1
$\begingroup$

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

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ 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, 2019 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, 2019 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, 2019 at 7:56

2 Answers 2

Reset to default
4
$\begingroup$

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.

Improve this answer
$\endgroup$
8
  • $\begingroup$ 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, 2019 at 6:30
  • $\begingroup$ Penguino, your solution for 8 is great and Jaap, I like your solution for 9. $\endgroup$
    – Ola Ström
    Apr 11, 2019 at 12:21
  • 2
    $\begingroup$ Ironically, 2 ? 2 ? 2 ? 2 ? 2 = 7 is easier  than 2 ? 2 ? 2 ? 2 = 7. $\endgroup$
    – Peregrine Rook
    Apr 14, 2019 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, 2019 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, 2019 at 11:04
3
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.