Measurement of Frequency Deviations in Equal Temperament Based on the Number of Humming Cycles

Problem Statement

Let $F$F be the frequency of the reference tone (the standard tone).

According to the tuning system known as “twelve-tone equal temperament,” in which the frequency ratios of all 12 semitones within an octave (the interval where the frequency exactly doubles) are set to be equal, the frequency increases uniformly by a factor of $2^{\frac{1}{12}}$2121​ for each semitone higher than the reference tone.

Let $f$f be the true natural frequency of a specific note that is three semitones higher than the reference tone. Based on the twelve-tone equal temperament, the true natural frequency $f$f can be expressed in terms of the reference tone’s frequency $F$F as follows:

When a specific note was played on an actual instrument, the tuning was slightly off, so the note sounded at a frequency $f'$f′ that was slightly lower than the true natural frequency $f$f. When this off-tune note was played simultaneously with the note at the true frequency $f$f, $n$n “beats” per second were observed. The number of beats per second is equal to the difference (in absolute value) between the frequencies of the two sound waves played simultaneously.

Here, if we let $\frac{f'}{F}$Ff′​ be the ratio of the actual frequency $f'$f′ of the offset sound to the reference frequency $F$F, the following equation is derived.

When the two terms on the right-hand side are simplified into a single fraction using the common denominator $F$F, the numerator takes the form $a \times 2^{\frac{1}{4}} - b$a×241​−b (where $a$a and $b$b are natural numbers). Determine the values of the coefficients $a$a and $b$b, and find the natural numbers that satisfy the following input format.

Constraints

• Frequency of the reference tone: $F = 440\text{ }\mathrm{Hz}$F=440 Hz
• Number of beats per second: $n = 3\text{ }\mathrm{s^{-1}}$n=3 s−1

Input Format

Enter the product of the two coefficients, $a \times b$a×b, as a natural number.