I actually came up with a different one: (75+(3*8))*5-4

For future reference, are calculators allowed? Pencil and paper? Can the intermediate answers and final answer be other than integers? For example, the closest whole-number response I found to Dead Cat’s challenge target of 477 was 475, but I found (with the help of a calculator) that [(5 x 2 - 1) + (4 ÷ 7)] x 50 equals approximately 478.571, which is a little closer. I know Giles’ got the exact answer, but would 478.571 be a legal response in the game?

I wrote a quick little app that checks all the permutations of the numbers recursively with different operands. For that problem it came up with 30ish solutions IIRC. It’s pretty quick but the search space is small (6! * 2^5). I retooled on a lark to see what targets were unreachable by a specific set of numbers. And some unreachable numbers do exist for some sets of working numbers.

On the TV show, the rules were (to the best of my recollection):

No calculators, but they had pencil and paper.

All intermediate and final answers must be integers.

Nice! I was thinking that most cases would end up with a problem where there is no exact solution, but you seem to have proved that the problems with no exact solution are rare. If I understand you correctly.

Tom Scud, you “won” the last challenge (you had the solution using the smallest amount of “working numbers”), would you care to propose one? Otherwise, if Tom Scud is away from the board, someone else can submit a challenge.

OK, here’s one.

working numbers 4, 8, 25, 100, 9, 2

target number 653

9 * ((100-25)-2) - 4

Tough one.

Ah, I was going to do a too-clever challenge with a target of 9, but I re-read the rules and there’s a minimum of 100.

Let’s say 1, 3, 5, 7, 50, 50

Target number 681.

I have no idea what if any solution there might be.

Missed edit window - make that

1, 2, 3, 5, 7, 50 and 681.

(50-1)*7*2-5=681

working: 1,3,4,6,9,10

value:379

((4 x 9) + 1) x 10 + 3 + 6 = 379

Countdown, mentioned by Dead Cat, is quite popular among students and other people with time on their hands on weekday afternoons, and the “numbers game” has been analysed quite a bit. Number games with all four big numbers are generally considered hardest, while the easy option is to choose one big number and five little ones. There are plenty of unsolvable games, and some with only one or two solutions. Some require you to use very big intermediate results, like this one: Numbers: 2, 4, 4, 6, 10, 50, Target: 687

But this one is not too hard:

Numbers: 3, 8, 8, 10, 75, 100

Target: 894

{edit} removed spoilery comment

(8 x 100) + (75 + 10 + 8) = 893

That depends on how big your numbers are. Over the domain of all numbers, puzzles with exact solutions are an infinitesimally small set compared to the ones without.

100 - 3 = 97

10 x 97 = 970

970 - 75 = 895

8/8 = 1

895 - 1 = 894

Target is 431, numbers are 1, 3, 4, 8, 10, and 25.

Target is 431, numbers are 1, 3, 4, 8, 10, and 25.

((8 * 10 + 25 + 3) * 4) - 1 = 431

Numbers:1, 3, 4, 10, 50, 100

Target: 713

That’s tricky. I can get 712:

((3 + 4) * (100 + 1)) + (50/10)

If we allow rounding, then it’s ((50 * 100) / (3 + 4)) - 1

I got 712 by the same method - no rounding allowed as per post #24. 713 does not look possible - I was trying to work something around 89 x 8 = 712 (and then adding the 1), but there aren’t enough numbers to make the 89 and the 8.

Alternatively, (10 + 4) x (50 + 1) = 714, but then you’ve lost the 1. In *Countdown*, it didn’t matter how many numbers you used, so this would have equal status with the 712 above.

Using a calculator, I discovered that 713 = 31 x 23, but I don’t think the right numbers are available to make these either - because 31 and 23 are both primes, you need at least 3 numbers to make each of them. I’d be interested to know whether or not 713 is possible.

It is solvable, I checked before I posted it.

ETA:But the solution is at home, and I’m at work.

You’re not supposed to do that!

Part of the fun is not knowing if the problem is solvable.

You can get 713 using a standard Vorderman split multiplication manoeuvre:

Observe that 50 x 4 x 3 + 100 + 10 = 710.

Then we can get to 713 by adding the 1 before the multiplication by 3 , as follows:

(50 x 4 + 1) x 3 + 100 + 10 = 713

Next target is 448

Numbers: 1, 5, 5, 7, 10, 25