Iron chain
A man has nine pieces of iron chain, with these sizes:
one 3-link chain
one 4-link chain
two 5-link chains
three 6-link chains
one 7-link chain
one 8-link chain
He wants to join them into one long 50-link chain, so he takes them to the hardware store. The store informs him that there is a $1 charge for cutting open any link, and a $2 charge for welding an open link back together. But he can buy a brand new 50-link iron chain for only $25. Is there a cheaper solution?
(This is based on a puzzle from a very old book by A. K. Dewdney. Now that I've finished writing it, I realize that $25 for a 50-link iron chain is insanely expensive. But go with it.)
posted by Ross at
12:52 PM
17 Comments:
$24 to get the work done. but i think i would just get the new one and have all the smaller chains if you need them in the future
Anon, there is a cheaper solution than that
$21 ... split the 7 link chain and use each link to join the others
$12
Split every link of the 4 link chain. The 4 links can then be used to join the 5 remaining sections.
Not quite. There are nine chains; if you split a seven link chain, you can only join eight pieces.
You're very close, though ... keep going
DualAspect, there are nine segments to begin with, not six
But if you use all 7 links of the 7 link chain ýou are only joing 7 lengths
ignore that - typing too fast
9 segments - splitting the 7 chain will join 8 lenghts, the 9th being the joiner of 7 joining lengths
Oh yeah, so there is. That'll teach me to read the question.
... 7 joining LINKS
Damn. You're right, Anon, I'm wrong.
The answer I got from the book said you had to break up both the 3-link and 4-link chains. But that's wrong; it ends up having an extra link -- which you can join onto the end but it isn't required.
Sigh.
I don't explain myself well at times ... I'm a Kiwi, more used to just doing it :-)
Ross. The book is also right. The single remaining link is needed, otherwise the chain wouldn't be 50 links, as required.
What I meant by "not required" is a bit nebulous, but here goes.
By using every link from the 7-link segment, every link is used to join two other segments.
If the 3-link and 4-link segments are broken up, only 6 of the 7 links end up joining two other segments. The 7th link just hangs on the end ....
----------------
Ah, wait, now I see what the book was talking about! The original problem wanted a CLOSED LOOP, not just a straight chain. With that additional requirement (which I unfortunately left out of my problem post), the 7-link segment is not sufficient. You have to use the 3-link and 4-link segments.
I had wondered about whether you had meant a closed chain, but you used the word "long", which I somehow took to mean in a straight line.
With the 3 and 4, you still have to pay $3 to attach the single link. So it makes no difference to the final chain or the price.
I myself misread the problem when reading the book, and in rephrasing I added the word "long" by mistake.
Right, it makes no difference to the price. You're still doing seven breaks and seven welds, for a total of $21.
In the original book problem, though, the only way to make a closed 50-link chain is to break up two segments, not just one.
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