Henry Ernest Dudeney/Puzzles and Curious Problems/347 - The Iron Chain/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $347$

The Iron Chain
Two pieces of iron chain were picked up on the battlefield.
What purpose they had originally served is not certain, and does not immediately concern us.
They were formed of circular iron links (all of the same size) out of metal half an inch thick.
One piece of chain was exactly $3$ feet long, and the other $22$ inches in length.
Now, as one piece contained exactly six links more than the other, how many links were there in each piece of chain?


Solution

The two pieces of chain contained $9$ and $15$ links respectively.


Proof

The difficult bit here is getting the picture.

In order to visualise the situation, we present a diagram of a $3$-link chain:

Dudeney-Puzzles-and-Curious-Problems-347-solution.png

We see that the length of a chain is equal to:

the total lengths of the inner diameters of each link

plus:

the thickness of the link for both of the links at either end.

Let $d$ inches be the inner diameter of one link.

Let $n$ be the number of links in the shorter chain.

We are given that the links are made of metal half an inch thick.

So twice the thickness of the links is $1$ inch.

Recall there are $12$ inches to the foot, so a $3$ foot length is $36$ inches

So, we have that:

\(\text {(1)}: \quad\) \(\ds 22\) \(=\) \(\ds n d + 1\)
\(\text {(2)}: \quad\) \(\ds 36\) \(=\) \(\ds \paren {n + 6} d + 1\)
\(\ds \leadsto \ \ \) \(\ds n d\) \(=\) \(\ds 21\)
\(\ds n d + 6 d\) \(=\) \(\ds 35\)
\(\ds \leadsto \ \ \) \(\ds 6 d\) \(=\) \(\ds 14\)
\(\ds \leadsto \ \ \) \(\ds d\) \(=\) \(\ds \dfrac {14} 6 = 2 \tfrac 1 3\)
\(\ds \leadsto \ \ \) \(\ds n\) \(=\) \(\ds 21 \div \dfrac {14} 6\)
\(\ds \) \(=\) \(\ds 9\)

So the shorter chain has $9$ links, while the longer chain has $9 + 6 = 15$ links.

$\blacksquare$


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