Question 207081
I don't know of the name either and I'm pretty sure that there isn't a formula. You just have to rearrange the values and try out different things.



Here are all the ones I could find (I couldn't find them all. So I'm not sure if they are all possible or not). I'm sure there are other ways to find them (as they are not unique)


<pre>

9*(5-6+1)    =   0 
5-1-9+6      =   1 
1*5-9+6      =   2 
1-9+5+6      =   3 
6-(1+9)/5    =   4
15/(9-6)     =   5  ... don't know if this is legal?
9*((5-1)/6)  =   6
9-1+5-6      =   7
1*9+5-6      =   8  
1+9+5-6      =   9
1*9-5+6      =   10  
1+9-5+6      =   11 
((9+1)/5)*6  =   12 ... the example given

5*(9-6)-1    =   14
5*(9-6/1)    =   15
5*(9-6)+1    =   16

9*(6-5+1)    =   18 
9-1+5+6      =   19
1*9+5+6      =   20 
1+9+5+6      =   21  
1-9+5*6      =   22 
6*(9-5)-1    =   23
6*(9-5*1)    =   24
6*(9-5)+1    =   25
5*(6+1)-9    =   26
6*(5+1)-9    =   27
(6+1)*(9-5)  =   28

6*(9-5+1)    =   30
</pre>


Note: for the question if it was legal or not, I thought about putting the digits 1 and 5 to get 15 (the number fifteen), but I wasn't sure if your teacher allowed it or not.