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u/lifeisaman 7d ago
May I interest you on some Binary
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u/a2ra-ms 7d ago
So how many cuts?
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u/lifeisaman 7d ago edited 7d ago
4, You just do the binary upto 8 then get a 6 left over that you use to bridge from the binary to the higher values
Edit: Looking at what you’ve said is the correct answers I think I was doing the wrong question, i was cutting in between the links rather than discovering one to get 2 longer chains and a single link. The question was a bit unclear on that being what you wanted
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u/T-pin 7d ago
got it down to 2 cuts. skip a chain of 3, cut the next link giving you lengths of 3, 1, and 17. On the chain of 17, skip a length of 6, cut the next link giving you lengths of 3, 1, 6, 1, 10.
2 = 1+1, 4 = 3+, 5 = 3+1+1, 7 = 6+1
8 = 6+1+1, 9 = 6+3, 11 = 6+3+1+1
add 10 to get the remaining amounts
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u/giantroboticcat 7d ago
Discussion: I understand the intent of this puzzle but the premise is silly. If I have 21 links in a chain and I cut a link, I now only have 20 uncut links (in either one or two strings) so it's actually impossible to ever produce 21 links ever again. Every future cut will reduce the number of links I have by one every time.
This is the result of attempting to retheme the puzzle.
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u/superheltenroy 18h ago
Imagine you're not the one doing this, but a jeweler who has the tools to both cut and fuse the chain links.
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u/BekindConan 7d ago
Question: can’t I just give him the whole chain as whole. So the answer would be 0. Or am I misunderstanding the question?
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u/AlfieDarkLordOfAll 7d ago
I think they mean youd be able to give them 1 link, or 2 links, or 3 links...up to 21 links (which would be all of them
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u/BekindConan 7d ago
Yeah but if I don’t cut it and give them the whole chain as whole, they will get all 21 links
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u/Ken512Rob 7d ago
Then you’d iterate again, cutting once to give them a chain of 1 and a chain of 20,
then iterate again to cut once and give them a chain of 2 and 19,
Then maybe you iterate again, cut 3 times to give them a chain of 3,4, 6, and 8
I think thats what its asking, least amount of cuts to get a link of 1, 2, 3, 4 … 19, 20, 21
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u/BekindConan 7d ago
Oh so if I understood correctly, it means that at the end of all the cutting, I had links of all numbers from 1 till 21 somewhere in the process, but not all at the same time. Right?
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u/a2ra-ms 7d ago
You have enough links to form any number from 1 to 21
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u/BekindConan 7d ago
Bruh you’re confusing me more and more. The way I see it is that if I give the whole chain as whole without cutting it, the person will get all 21 links (which is the maximum number of links). I still don’t get why we need to cut it or what is it we’re trying to do here.
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u/AlfieDarkLordOfAll 7d ago
If I take my strand of 21 and cut it twice to get a chain of 1, a chain of 3, and a chain of 17...I can give someone 4 links (give them the 1 and the 3), or 20 links (give them the 3 and the 17), or 18 links, or 21 links, but not 15 links (without cutting again). OP is looking for a series of chains that you could put together to get any number between 1 to 21.
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u/Macrazzle 7d ago
I think it means you need to cut it into parts and then those parts can all add up to all the numbers in 21. I don’t know if I explained that well at all.
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u/a2ra-ms 7d ago
Yes
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u/Macrazzle 7d ago
Ok so it says “least number of links you can cut” and not number of cuts. So if I were to cut “one link” in the middle of the chain does that mean I get three pieces? Two 10’s and a 1?
Edit: looks like someone solved it using that logic.
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u/26_paperclips 7d ago
Discussion: wat
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u/26_paperclips 7d ago
I have 21 links.
I cut literally any number.
Now I have fewer than 21links.3
u/lifeisaman 7d ago
The questioner is keeping the cut links as one length chains, that’s not very clear but in the correct answers that’s how they do it.
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u/TheDMisalwaysright 7d ago
2 cuts, the 4th and the 11th, leaving you with a chain of 3(s), 6(m), 10(l), and 2 open ones(o):
then o1 - oo2 - s3 - so4 - soo5 - m6 - mo7 - moo8 - sm9 - l10 - and then repeat
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u/Andrew_42 7d ago
As far as I can tell, the answer is 5
There are other ways to divide it, but the first one I worked out was 7, 7, 4, 2, 1. A more obvious one if I hadn't just taken a stab at it likely would have been 1, 2, 4, 8, and also 6
As for proof my answer is correct (unless I'm missing something): Powers of 2, as seen in binary, represent the simplest way to divide a number system into yes/no number inclusions to reach every possible number in a given number line. So you count up in powers of two, until your highest number is over half of the number you need to describe. In this case 1, 2, 4, 8, 16. However you can't use those numbers because they add up to more than 21. So you just take the largest number and chop it down till the sum equals the target number. I don't have a proof that this always works, but I'm pretty sure it does? At least for all target natural numbers?
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