# MATH PROBLEMS?

Maths in the City posted this on twitter:

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

Multiply out the first one:

$\left(10x\right)\left(13i{\right)}^{2}=-1690x$

So $x$ needs to be of the form

$-\frac{n}{1690},$

Now, to look at the second part!

Substituting in the expression we had for $x$, we get

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

Let’s rearrange that to find $y$ in terms of $m$ and $n$:

$\begin{array}{rl}\frac{-2.362mn}{1690}& =\mathrm{sin}\left(\frac{-ny}{1690}\right)\\ {\mathrm{sin}}^{-1}\left(\frac{-2.362mn}{1690}\right)& =-\frac{ny}{1690}\end{array}$

At this point, note that $\mathrm{sin}\left(-z\right)=-\mathrm{sin}\left(z\right).$

$\begin{array}{rl}\frac{ny}{1690}& ={\mathrm{sin}}^{-1}\left(\frac{2.362mn}{1690}\right)\\ \\ \\ y& =\left(\frac{1690}{n}\right){\mathrm{sin}}^{-1}\left(\frac{2.362mn}{1690}\right)\end{array}$

Which is great!

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

$0\le \frac{2.362mn}{1690}\le 1$

i.e.

$mn\le \frac{1690}{2.362},$

or,

$mn\le 715.495343.$

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

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