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Maths in the City posted this on twitter:

In order to make a number we can call, we need both of \[n=(10x)(13i^2)\] and \[m=\frac{\sin(xy)}{2.362x}\] to be integers.

Multiply out the first one:

\[(10x)(13i)^2 = -1690x \]

So \(x\) needs to be of the form


Now, to look at the second part!

Substituting in the expression we had for \(x\), we get

\[ m = \frac{\sin \left( \left(\frac{-n}{1690} \right) y \right)}{2.362 \left(\frac{-n}{1690} \right)}. \]

Let’s rearrange that to find \(y\) in terms of \(m\) and \(n\):

\[ \begin{align} \frac{-2.362mn}{1690}&= \sin \left( \frac{-ny}{1690} \right) \\ \sin^{-1}\left(\frac{-2.362mn}{1690} \right) &= -\frac{ny}{1690}\end{align} \]

At this point, note that \(\sin(-z) = -\sin(z).\)

\[ \begin{align} \frac{ny}{1690} &= \sin^{-1}\left( \frac{2.362mn}{1690}\right) \\ \\ \\ y &= \left( \frac{1690}{n} \right) \sin^{-1}\left( \frac{2.362mn}{1690}\right)\end{align} \]

Which is great!

Since \(\sin^{-1}\) is defined only on values in the range \([-1,1]\), and we want both \(m\) and \(n\) to be positive, we need

\[ 0 \leq \frac{2.362mn}{1690} \leq 1 \]


\[ mn \leq \frac{1690}{2.362}, \]


\[ mn \leq 715.495343. \]

So if you have “math problems”, call any number 1-800-n-m such that \(mn \leq 715\).



I’ll take your word on this. If any chance of me solving, I would have tried, but forget Trig 100%, so not a chance …

Sure that the x in the last term (2.362x) is supposed to be in the denominator?

$i$ is the square root of $-1$, i.e. $i^2 = -1$, so it’s the minus sign.