# Problem: Measuring Time with String and Match

Play with your problems before you try and solve them. Allow me to demonstrate with the following little brainteaser.

Problem 1. You have two lengths of string and as many matches as you like. When you light one end of a single length of string, it takes one hour to burn the full length. Can you devise a way to measure out 45 minutes?

My immediate reaction is why-on-earth-would-I-want-to but being a mathematician I have long since learnt that sometimes we put how before why.

Let’s forget the problem and think about the situation. Put yourselves in the shoes of this bored pyromaniac. You’ve got some string and some matches. All you can really do is light things up.

Try and picture all of this in your head. Imagine lighting the end of a piece of string. You sit there and watch as it burns from one end to the other. Then you try and pinpoint the moment that your life became so dire that you had a spare hour to watch a piece of string burn.

In this mad sandbox we have in our head, what else can we do but light string? Have we not done everything we can do?

Well, there is one other thing: we can light both ends of a single piece of string. And if we do this, how long will that string take to burn down?

Now, I leave solving the actual problem to you. I just wanted to help your imagination along a little bit.

Once you have solved Problem 1, you may try something a bit fiercer viz. the following two problems:

Problem 2. Given $N$ pieces of string, each of which take one hour to burn if a single end is lit, what possible lengths of time can you measure out?

Problem 3. What is the minimum number of pieces of string required to measure out 27 minutes?

1. Sarthak says:

There are two observations that I could come up with (As a crazy Pyromaniac of course!). The first one is as you mentioned that one could light up a singular length of string from both ends, and they would meet somewhere, length covered by each spark of fire would vary yet the total distance would be say $latex l$ but they would take the same time to traverse those individual distances that add up to $latex l$ with the length of the string corresponding to an hour therefore the we can measure 30 mins doing that. The second thought that I had was what if I light up the opposite ends of both the strings! they would have to meet somewhere not physically but concurrently (I would like to think of this as the IVT for strings), and then quickly realized that wherever the meet concurrently also keeps a measure of time thus piecing both the components together one come to the realizations that “Drumroll”:

Light the opposite ends of the first string and light up the second string any end will do, this keeps tracks of time 30 mins. However to measure the extra 15 minutes (half of 30 minutes), we now light the unlighted end of the second string, and when the dust settles you should have your 45 minutes measured.

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• Adrian Dudek says:

Nice work Sarthak! Sounds about right to me?

How did you go with the other problems? I should have been clearer around Problem 3 (and perhaps I will still edit the post) but the goal is not to measure out 27 minutes from the lighting of the first match (as this is impossible!) but just any exact 27 minute period.

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• Sarthak says:

In the entire argument some proper fractions may not be ‘obtainable’.

In a preliminary reading n-strings case reminds of defining sums on linear vector spaces (Cartesian sum) by which I mean numbers of the form $\frac{1}{2^k}$ (cause that more or less what we are doing ). Moreover, lets say the ordered set of measurable time for n-strings is $T_{n}$. This subsequently affects $T_{n+1}$, we start with $T_{1}=\big\{1,1\2\big\}$ the next leads us to $T_2=\big{1,1/2,1/4,1+1/4\sim 3/4\big}$. My conjecture would be for the second problem that all the elements in the set $\big{\frac{k}{2^n}: 1\le k\le 2^{n}-1\big}$ are ‘measurable’, well this checks out for at the very least the 3-string. Therefore we can prove the above my induction. But I think one can also think of getting $T_{n+1}$ from $T_{n}$ by simply copying all the elements from the latter and then also including elements that are $\sum_{i \ne j}i+j:i,j\in T_{n}$?

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2. Sarthak says:

The above didn’t render and is a draft(would be good if the earlier message can be deleted altogether) :

In a preliminary reading n-strings case reminds of defining sums on linear vector spaces (Cartesian sum). Moreover, lets say the ordered set of ‘measurable’ time for n-strings is T_{n}. This subsequently affects T_{n+1}, we start with $T_{1}=\big\{0,1,1\2\big\}$ the next leads us to $T_2=\big\{0,1,\frac{1}{2},\frac{1}{4},1+1/4\sim 3/4\big\}$. My conjecture would be for the second problem that all the elements in the set $\big{\frac{k}{2^n}: 1\le k\le 2^{n}-1\big}$ are ‘measurable’, well this checks out for at the very least the 3-string. Therefore we can prove the above by induction. But I think one can also think of getting $T_{n+1}$ from $T_{n}$ by simply $T_{n+1}=\sum_{i \ne j}(i+j)/2:i,j\in T_{n} \cup T_{n}$. Therefore we can measure time of the form $m+k/2^n$ where $m$ is some integer. I think the for the former the union with $T_n$ is may not be needed.

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3. Sarthak says:

Haha it doesn’t render above as well. Well the set in the conjecture was k/2^n where 1<=k<=2^n -1. And T_1={0,1,1/2}.

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