## MATH PROBLEMS?

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

\[-\frac{n}{1690},\]

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

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

i.e.

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

or,

\[ mn \leq 715.495343. \]

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 …

2013-09-09 18:16:52

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

2013-09-11 08:45:09

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

2013-09-11 08:54:59

For the first multiplication, what happened to i?

2014-03-24 05:36:08

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

2014-03-25 13:27:52