TITLE: Can you complete the expression $2 \underline{ }\, \underline{ }\, \underline{ } \,\underline{ } 5 = 2015$?
QUESTION [20 upvotes]: Can you complete the expression
$2 \underline{ } \, \underline{ }\, \underline{ } \, \underline{ } 5 = 2015$
and make it correct by replacing two underscores with a selection of the operational symbols $+, - , /, \times$ and the other two underscores with digits $0,1,\ldots,9$?

I have been working on this problem for quite a while now where my main strategy has simply been trial and error. However, I still can't seem to find a combination of operational symbols and digits where the result gives me 2015. If this is in fact not possible I would greatly appreciate an explanation and if it is possible, I would greatly appreciate an explanation of how you were able to solve it.
Thank you.

REPLY [5 votes]: Brute force solution
I originally wrote the solution for 5 slots, with only 4 slots this approach proves that only two solution candidates appear: 2010+5 and 2020-5. Obviously these don't match the criterium so it is proved that no solution exists. For illustration I will hereby show how I found the solution for when there are 5 slots.
Approach
It can easily be seen that there there are just a limited number of possibilities here. A simple upper bound can be found by considering that each slot can be occupied by 14 different characters, and there are 5 slots. As such we can simply try roughly half a million possibilities (14^5 to be exact) and evaluate the solution.
Matlab code
v = ['0':'9' '+' '-' '*' '/'];

results = [];
for a=v
    for b=v
        for c=v
            for d=v
                for e=v
                    x = false;
                    s = ['x = 2' a b c d e '5 == 2015;'];
                    try 
                        eval(s);
                    catch
                    end
                    if x
                        results = [results; s];
                    end
                end
            end
        end
    end
end

Results
x = 2000+15 == 2015;
x = 2010+05 == 2015;
x = 2010++5 == 2015;
x = 2010--5 == 2015;
x = 2020+-5 == 2015;
x = 2020-05 == 2015;
x = 2020-+5 == 2015;
x = 2030-15 == 2015;
x = 2040-25 == 2015;
x = 2050-35 == 2015;
x = 2060-45 == 2015;
x = 2070-55 == 2015;
x = 2080-65 == 2015;
x = 2090-75 == 2015;
x = 20+1995 == 2015;
x = 2100-85 == 2015;
x = 2110-95 == 2015;

Conclusion
Most combinations do not meet the restriction of having exactly 2 operators, but some do. My personal favorite which I would personally consider to be the answer:
2010--5 == 2015