Maths in the City posted this on twitter:

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

Multiply out the first one:

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

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 $\mathrm{sin}(-z)=-\mathrm{sin}(z).$

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

Which is great!

Since ${\mathrm{sin}}^{-1}$ is defined only on values in the range $[-1,1]$, 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 $mn\le 715$.

## Comments

## Comments

rose82

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 …

Alex

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

Christian Perfect

Yeah, pretty sure. Why wouldn’t it be?

I have no idea.

For the first multiplication, what happened to i?

Christian Perfect