Originally posted by SwissGambit
Given the equation
123456789 = 2002place + or * (plus or multiply) symbols as needed between the digits on the left side of the equation such that the equation is true.
Check me if wrong:
I believe there are 2^8 possible arrangements of * and + between the 9 numerals. Call these the *+ arrangements. Example: **+*++**. That is 256 testable arrangements. Certain *+ arrangements can be excluded, such as any group of 8 that ends with ***, because 6*7*8*9>2002, no matter how 12345 are operated upon. It can also be noted that by the rules of math, multiplication supervenes upon addition, such that a*b+c*d works out as (a*b)+(c*d), not as a*(b+c)*d.
Firefox allows such numeric expressions to be put in the search bar and with an = sign, will deliver the result. It uses the above math rule about * and +.
I see no way to avoid a tedious brute force approach other than that there are some exclusions available like the above.
We are working in base 10, right?
Is there any significance to the arrangement that yields 2002? Or could any of the 256 possibilities have been presented? It's an easy puzzle to construct, if the latter.
Related question: is there only one *+ arrangement that yields 2002?
Can the general conjecture that no two *+ arrangements yield the same number, be proven or disproven, using a non-brute force method of proof?
It would be easier to work on a smaller set, like 1234. But do the results extrapolate?
I find these questions to be more interesting than the puzzle presented, but they wouldn't come up without the puzzle being presented, so thanks.