Chain links. Given a chain with 7 links, what is the fewest number of links that need
to be cut so that you could hand a person any number of links from 1 to 7? A cut link is
counted as a link (weld it back together after you remove it).
Repeat with 23 links. Repeat with 63 links. What formula relates n, the number of cut
links, to the maximum length of the chain?
Chain of Seven Links:
Given a chain of 7 links, 1 is the fewest number of cuts that need to be made in order to
create any number 1-7. The cuts would be made such that there would be a single link,
two chain link, and a four chain link.
1
2
3
4
5
6
7
Chain of 23 Links:
Given a chain of 23 links, 2 is the fewest number of cuts that would need to be made in
order to create the numbers 1-23. The cuts would have to be made such that there
would be 2 single chain links, a three chain link, a six chain links, and a 12 chain link.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
Chain of 63 Links:
Given a chain of 63 links, 3 is the fewest number of cuts that would need to be made in
order to create the numbers 1-63. The cuts would be made in such a way that there
would be three single chain links, a four chain link, an eight chain link, a sixteen chain
link, and a thirty-two chain link.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
Generalization:
To help us generalize a pattern, we made a table of the data.
Cuts Minimum Number of Links Maximum Number of Links
1
2
7
(1 + 2 + 4)
2
8
3
24
4
64
5
160
n
n(2n-1 + 2n-1 – 1) + n =
n(2n-1)+n
23
(1 + 1 + 3 + 6 + 12)
63
(1 + 1 + 1 + 4 + 8 + 16 + 32)
159
(1 + 1 + 1+ 1 + 5 + 10 + 20 + 40 + 80)
383
(1 + 1 + 1 + 1 + 1 + 6 + 12 + 24 + 48 + 96 + 192)
(n + 1)(2n + 2n – 1) + n =
(n + 1)(2n+1 – 1) + n
We noticed that each time there were the same number of single links as the number of
cuts. This stands to reason because every cut can produce a single link. We also
noticed that the first number of links in the non-cut chain is the number of single links
plus one. After that the number of links doubles the number before it in the sequence.
This allowed us to come up with the generalized formula that for any number of links
between n(2n-1 + 2n-1 – 1) + n and (n + 1)(2n + 2n – 1) + n, the chain will require n cuts.
These formulas can be further simplified to n(2n-1)+n and (n+1)(2n+1-1)+n. Finally, we
needed to check the formulas to make sure they worked for the numbers that we found
through looking at data.
Cuts
(n)
1
2
3
4
Known
Min.
2
8
24
64
Formula Minimum
Known
Formula Maximum
n(2n – 1) + n
Max.
(n+1)(2n+1 – 1) + n
1(21 – 1) + 1 = 1 + 1 = 2
7
(1+1)(21+1 – 1) + 1 = 2(3)+1=7
2(22 – 1) + 2 = 6 + 2 = 8
23
(2+1)(22+1 – 1)+2=3(7)+2=23
3(23 – 1) + 3 = 21 + 3 = 24 63
(3+1)(23+1 – 1)+3=4(15)+3=63
4(24 – 1) + 4 = 60 + 4 = 64 159
(4+1)(24+1 – 1)+4=5(31)+4=159
5(25 – 1) + 5 = 155+5 =
5
160
383
(5+1)(25+1 – 1)+5=6(63)+5=383
160
The formulas work with both the data and the reasoning for developing them, so we
can be fairly certain that they are correct.