The pitch of a musical note is determined by its frequency, with higher notes corresponding to higher frequencies and lower notes to lower frequencies. For example, the frequency of a tuning fork is exactly 440 Hz, and this corresponds to the one-line A note. Intervals which sound pleasant to the human ear correspond to nice ratios of frequencies: for example, if two notes are an octave apart, the frequency of the higher note is exactly twice the frequency of the lower note, so an octave corresponds to the ratio 2:1. The perfect fifth similarly corresponds to the ratio 3:2.

In Western music, an octave is divided into twelve notes: you can see the corresponding twelve keys of the piano below. But why exactly twelve? Could we make a piano with more or fewer keys per octave?

The answer lies in the mathematics of intervals. Starting on any note, and moving up or down by an octave or a perfect fifth can be described in terms of frequencies as follows:

• Move 1: Double the frequency of the note (move up on octave)
• Move 2: Halve the frequency of the note (move down on octave)
• Move 3: Multiply the frequency of the note by $3/2$ (move up a perfect fifth)
• Move 4: Multiply the frequency of the note by $2/3$ (move down a perfect fifth).

So on the piano, starting at the frequency of any given key (for example at 440 Hz), we should be able to apply any sequence of moves from the above and always land on a frequency of a key. But there are in fact infinitely many frequencies strictly between 440 Hz and 880 Hz that you could reach by applying some sequence of the moves above, which would require a piano with infinitely many keys in a single octave. So instead, in practice, we pick a number $y$ very close to $1$, and replace moves 3 and 4 with

• Move 3’: multiply the frequency of the note by $3y/2$ (move up an approximate fifth)
• Move 4’: multiply the frequency of the note by $2/3y$ (move down an approximate fifth).

If $y$ is close enough to $1$, most people can't hear the difference between moves 3 and 3’ or 4 and 4’. Pianos are tuned choosing $y \approx 0.99887$, which results in exactly eleven numbers strictly between 440 Hz and 880 Hz that can be reached by applying the moves 1, 2, 3’ and 4’, hence the twelve keys in an octave.

We could have chosen $y$ slightly closer to $1$, at the cost of having more notes in an octave. What is the lowest possible number of keys in an octave where $y$ is chosen to be closer to $1$ than the approximation giving twelve keys?

Remark: in practice, the choice of $y$ for pianos also takes other intervals – such as the thirds and the sixths – into account, but for the sake of the problem we simplified things.

This problem was first solved on Wed 5th February at 4:09:24pm

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