Shouldn't this go under the Math Challenge Problems section?
So you have 6 numbers, ranging from 1-24. Five of the six numbers must equal the 6th number by using any of the 4 operations (+, -, *, /) and using each number only once.
So for example:
3 7 12 19 21 | 5
So you need to get the number 5 by moving around and using the operations on those five first numbers.
Now, find an example where this is impossible!
EDIT: I think I found one:
5 11 17 19 23 -> 1
See if you can find away this would work, cause I cannot!
Also, how would one go about proving it is a valid counter example?
For some reason in my head I was thinking you can only add or subtract a 1, because multiplication or division by 1 were pointless Nevermind.
I don't think 24 having too many factors is a fair reason to rule it out though, if you pose the same problem but with 4 numbers instead of 5, then there are plenty of combinations of 4 numbers that can never be made to equal 24.
(17 + (19 + 11) / 5) / 23 = 1
17 / ((19 - 11) * 5 - 23) = 1
((23 - 17) * 5 - 11) / 19 = 1
(11 + (23 + 17) / 5) / 19 = 1
11 / ((23 - 17) * 5 - 19) = 1
(19 - 11) * 5 / (23 + 17) = 1
(23 - 17) * 5 / (19 + 11) = 1
(19 + 11) / (23 - 17) / 5 = 1
((19 - 11) * 5 - 23) / 17 = 1
(23 - (19 + 11) / 5) / 17 = 1
23 / ((19 - 11) * 5 - 17) = 1
11 / (19 - (23 + 17) / 5) = 1
19 / (11 + (23 + 17) / 5) = 1
(23 - 11) / (19 - 17) - 5 = 1
17 / (23 - (19 + 11) / 5) = 1
23 / (17 + (19 + 11) / 5) = 1
I have written a program that can check them.
As for proving a counter example or proving there are none, that's rather difficult I think.
The only way I can think of showing that a solution is true is by an exhaustive algorithm. If someone were to say they have a solution, I can verify it computationally using a program that exhausts all possible parsings of the numbers and symbols, and shows none of them equal the target number.
Whether or not you see this as a proof is your call, you'd have to have confidence that my program really did check all possibilities.
True there aren't many ways of organizing the +-/* and brackets with 4 numbers, and there are only 24C5=42504 ways to exhaust all possibilities. It's probably only going to take a few hours to check all of them with a not so efficient program. But I don't think computer provided exhaustion is really a proof...
But we're using + - / * on 5 numbers, with a 6th as a target. There are so many combinations.
Why wouldn't you see this as proof? Do you think the 4 colour theorem proof is not really a proof? It's not the most elegant of proofs, actually, it's probably the most inelegant proof I've ever heard of, but it's proof nonetheless.