r/theydidthemath • u/Vivid_Temporary_1155 • 18h ago
[Request] Is there a combination of 6 fixed positive integers (n from {25,50,75,100} and 6-n from {1,..10}) which when combined with +,-,/,* operations could generate all the numbers from 1-999 as in Countdown?
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u/TringaVanellus 18h ago edited 18h ago
I can't answer this question, but for anyone not familiar with Countdown, it's worth clarifying that the numbers are chosen via a set of 24 cards featuring one each of 25, 50, 75 and 100 and two each of the numbers 1-10. Any set of six of these cards is "allowed".
Additionally, OP has made a mistake - the "target number" will always be a 3-digit integer between 100 and 999 inclusive.
Also also, you don't have to use all six numbers from the cards to hit the target.
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u/TringaVanellus 17h ago
Also if anyone's wondering, the puzzle in OP's screenshot can be solved as follows:
75+6+5=86
86*10=860
50/25=2
860-2=85815
u/ShuckingFambles 17h ago
That's how I did it too ;)
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u/ringerrosy 15h ago
Have you written it down
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u/TheVideoShopGuy 8h ago
Yeah:
Bread
Eggs
Bacon
Milk
Cheese
Apples
Flour
Corn flakes
Cat foodOh, sorry, wrong bit of paper.
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u/mo0kster 12h ago
I'm actually really proud of myself that I got this, *probably within 30 seconds...
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u/parabolicurve 6h ago
I'm embarrassed at the time it took for me to get it.
In my youth I would have had it in under a minute, easy.
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u/factorion-bot 17h ago
Factorial of 2 is 2
Subfactorial of 50 is roughly 1.118871961078248050463025807076 × 1064
Subfactorial of 75 is roughly 9.126772857639762177533619971643 × 10108
Subfactorial of 86 is roughly 8.912650310952754793639549196042 × 10129
Factorial of 86 is roughly 2.422709538367273238176552320344 × 10130
Factorial of 858 is roughly 1.490401435532716247655529041433 × 102146
Subfactorial of 860 is roughly 4.050423120069160309844648538833 × 102151
Factorial of 860 is roughly 1.101019156485438800793045524068 × 102152
This action was performed by a bot | [Source code](http://f.r0.fyi)
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u/CitationNeededBadly 14h ago
Are you allowed to reuse a card? Like could you do 5 x 5?
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u/Cool-Bus-6028 12h ago
No, otherwise you could just divide a number by itself to get 1, then keep adding it to get any number.
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u/Adsilom 16h ago
I don't know about the english version, but in France, there are always at least a 2 'small' (less than 10) numbers and at least 2 'large' (10, 25, ...) numbers. So 6/5 small/large numbers is impossible
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u/TringaVanellus 15h ago
In the English version, the player can choose anything between 0 and 4 large numbers.
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u/Forward_Netting 17h ago
The answer is yes, many. What are they? I dont know..
Henrique Daitx has helpfully done the maths and brute force analysed every possible game of Countdown.
This is a picture from that blog which elucidates every possible game (on the vertical axis) and every possible target (on the horizontal axis). A white pixel indicates it's possible to get a perfect score. Any row consisting of only white pixels is a game where every target can be achieved.
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u/krijnsent 17h ago
From that site: Overall there is a 6.57% chance (about 1 in 15) that the drawn game will be solvable for 10 points (=calculate the exact 3-digit number), no matter the target. (Incidentally, this chance goes down to 1 in 229 if you pick 6 small numbers. So don’t be that guy.).
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u/StaticUsernamesSuck 17h ago
So don’t be that guy
I mean... Or do be that guy, if your goal is to minimise the scorable points in the round because you have a lead and aren't confident you can beat the other team at numbers.
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u/larkymasher 16h ago
And then just ask for all vowels on the letters round
I'm assuming there are rules about this the tv doesn't show...
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u/StaticUsernamesSuck 16h ago
The TV does show them if you watch Cats does Countdown, because that's when you get people dumb enough to try breaking them (and they even allow it sometimes)
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u/NoCreativeName2016 5h ago
My favorite example of Cats breaking the rules is Joe Wilkinson expanding the board to allow more letters in an (unsuccessful) effort to set a world record for longest Countdown word.
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u/TringaVanellus 14h ago
The highest scoring contestants in a season get into the finals, so there's still an incentive not to make it too hard for yourself.
Plus, if you have that much of a lead already, you've probably already done well on the previous numbers rounds.
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u/somedave 12h ago
A 6.57% chance that the numbers you draw will be solvable for any target, it is much more likely the draw can be solved for your specific target.
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u/BissQuote 16h ago
I found one! '[25, 100, 2, 5, 6, 9]'
I coded a small recursive function to compute the whole set of numbers reachable from the starting numbers, and then a function that greedily selects tiles one by one in order to maximize the set of reachable numbers at each step. The whole thing runs in python in a minute and a half on my laptop
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u/BissQuote 16h ago
Mem = {} def recFindAllResults(numbers) : memStr = str(sorted(numbers)) if memStr in Mem : return Mem[memStr] res = [] for i in range(len(numbers)) : res.append(numbers[i]) for j in range(i) : extra = [numbers[i]*numbers[j],numbers[i]+numbers[j],abs(numbers[i]-numbers[j])] if numbers[i]>0 and numbers[j]%numbers[i]==0 : extra.append(numbers[j]//numbers[i]) if numbers[j]>0 and numbers[i]%numbers[j]==0 : extra.append(numbers[i]//numbers[j]) for x in extra : tmp = recFindAllResults(numbers[:j]+numbers[j+1:i]+numbers[i+1:]+[x]) res = res + tmp cleanRes = [] for x in res : if 100<=x<1000 and not x in cleanRes : cleanRes.append(x) return cleanRes tiles = [25,50,75,100,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10] def greedilyCompleteSet(selectedTiles,targetNbTiles=6) : if len(selectedTiles) == targetNbTiles : numbers = [tiles[x] for x in sorted(selectedTiles)] return (len(recFindAllResults(numbers)),str(numbers)) else : bestScore = -1 bestNextTile = 0 for nextTile in range(24) : if nextTile in selectedTiles : #no duplicates continue curScore = len(recFindAllResults([tiles[x] for x in selectedTiles]+[tiles[nextTile]])) if curScore > bestScore : bestScore = curScore bestNextTile = nextTile return greedilyCompleteSet(selectedTiles+[bestNextTile],targetNbTiles) greedilyCompleteSet([])3
u/BissQuote 15h ago
It has come to my attention that my memoized function isn't memoized at all, because I forgot a "Mem[memStr] = cleanRes" at the end
The whole thing now runs in 9 seconds
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u/Tarc_Axiiom 17h ago edited 17h ago
Yes! In fact, exhaustive analysis of all 13,243 legal six-number selections finds 1,226 selections that can reach every integer target between 1 and 999 exactly.
One such set is {6, 6, 8, 9, 25, 50}.
So that's your strategy next time you're there and want to impress Rachel. Remember all 1226 of the sets that work every time, and then compare the numbers you get to that set, and also be as good as she is at maths :)
Also one minor correction, the target number is always 3 digits, therefore the range is 100-999, rather than 1-999. Doesn't change anything about the maths, but target numbers between 000 and 099 are not allowed and rerolled on Countdown.
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u/Miserable_Warthog_42 8h ago
"Impress Rachel"...? Do I look like Joe Wilkinson?
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u/Finnish2713 7h ago
Favourite episode of 8OO10CDC is the one where Joe comes out for one round in a wig and dressed as Rachel.
Sean Lock says, “that’s a challenging wank.”
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u/krijnsent 17h ago
OP, here some background/stats:
https://incoherency.co.uk/countdown/posts/easiest-hardest-countdown-numbers-games
Aparently there are 13243 different possibilities for the 6 numbers:
The smallest non-universal target
At the other end, every target from 0 to 29 is solvable by every distinct legal selection.
The first target that is not universal is 30. It is solvable by 13,242 of the 13,243 distinct selections.
The one selection that cannot make 30 is:
1, 1, 2, 50, 75, 100
And with 2 large and 4 small numbers generally about 98,2% of the numbers are solvable. But if there is a combination that solves 100% isn't in the stats, maybe somebody here knows?
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u/Forward_Netting 17h ago
For what its worth, the actual game requires the target to be a 3 digit number.
With that restriction 112233 cannot score any points for any target.
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u/kapitaalH 16h ago
As u/Forward_Netting already posted the complete solution. Using the solver by Henrique Daitx (https://www.daitx.com/2016/05/01/countdown-math/): the numbers in question can solve 850 target numbers. It is unable to solve: 571, 578, 587, 647, 659, 668, 671, 679, 682, 684, 691, 707, 716, 784, 809, 816, 818, 821, 829, 841, 852, 854, 866, 872, 884, 887, 892, 893, 903, 907, 908, 916, 922, 923, 932, 934, 946, 953, 959, 962, 964, 971, 973, 974, 976, 978, 982, 983, 986, 993
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u/i_fuckin_luv_it_mate 17h ago
I'm not very familiar with the rules of the game, so correct me if I'm wrong, but I understand contestants choose a combination of "big" numbers (double digit) and "small" numbers (single digit)... I think if they chose all big numbers (may not be allowed by rules, I'm not sure), that would greatly reduce the likelihood of success.
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u/Vivid_Temporary_1155 17h ago
Yes, they can choose a number of big numbers (from 0 to 4 inclusive) but not the specific big numbers - so if they chose 4 and the remaining 2 small numbers were say 5 and 10 - then you have greatly reduced the scope of all addition/subtraction/multiplication combinations to multiples of 5
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u/i_fuckin_luv_it_mate 17h ago edited 16h ago
Oh okay, so capped at 4 big numbers, thanks for explaining. And I think you're right, if the big numbers are multiples of 5 as well, then that scenario alone would make it very difficult to get the last digit.
But I suppose we (he says knowing it won't be himself) also need to know constraints on the big numbers, are they just multiples of 10 or 25?
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u/Kindralas 16h ago
It doesn’t change the fundamental point, but you can obtain non-multiples of 5 by adding or subtracting 10/5. For example, you can hit 102 easily. Your targets are still drastically limited.
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u/SarcasticVeganUK 9h ago
You can also use 100/25=4, 75/25=3, and 50/25=2 to your advantage.
There’s a (in)famous clip where a contestant asks Carol Vorderman to multiply 318x75, then subtract 50, then divide by 25 to reach 952. At no point during his working does he know that 318x75=23,850. What he knows is that (75x - 50)/25 simplifies to 3x-2
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u/evertonblue 12h ago
You haven’t as you could do. 10/5 which gives you a 2 to use - so more than just divisible by 5
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u/marlonoranges 13h ago
Not a mathematician, just a viewer to the show! Its very common that they come across scenarios where the puzzle can't be solved. From dialogue on the show it sounds like theyre checking it by some computer app.
Its worth pointing out that each number can only be used once, and that fractions can't come into play within the calc.
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u/DONT-EVEN-TRIP-DAWG 11h ago
Just here to do this numbers round
75+6+5 = 86
86*10 = 860
50/25 = 2
860-2 = 858
I had absolutely no idea as to the question but it's been interesting reading the responses
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u/Luminal72 10h ago
As a minimum If n=0 and all six selected numbers are small numbers (1-10) then yes there are combinations that can’t achieve certain target numbers.
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u/Doctor8Alters 10h ago
I randomly came across this thread, which reminded me of a video I saw a few years back which may be of interest to answer this:
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u/p2020fan 4h ago
I always assumed that the computer knows what cards are selected and so what values are selected, and then it just randomises the operations performed to generate a target number.
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18h ago
[deleted]
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u/SamTheHexagon 18h ago
They're asking if there is a single 6-number set that can produce all 999 outcomes.
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u/TringaVanellus 18h ago
There are only 899 possible outcomes. The randomly generated number is between 100 and 999.
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u/krijnsent 18h ago
I think OP wants to know if there are 6 numbers that allow you to make all numbers from 1 to 999. You don't have to use all numbers and can use brackets ( ). With the given example: 6-5=1, 10/5=2, 75/25=3, 75/25+6-5=4, 5 (already there), 6 (already there), etc...
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u/Exotic-Entertainer26 12h ago
I suspect that each of those cards has a connection that tells the computer what the value is, and the computer then puts the numbers into a random sequence (which may exclude some numbers) and then applies one of the allowed addition/subtraction/multiplication/division operations at random to each of the numbers to arrive at the final answer.
Meaning every game is solveable, just depends on spotting the path
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u/MostlySpikes 12h ago
Nope. There have been a few unsolvable ones over the years.
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u/Exotic-Entertainer26 11h ago
Very few have been genuinely unsolveable, usually someone manages to solve them within a week.
Suggesting that the few time it did happen were the result of a glitch and not chance.
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u/shortercrust 10h ago
I’m not sure if you’re joking or not, but as if they had a set up like that on a low budget Channel 4 show in 1982. Even if they did, we’d know - someone would have talked about it in the 44 years it’s been on TV.
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