Pick a Number ...
… any number.
You’ll probably want this number to have three or four digits. Fewer might be boring and more may get a bit complicated.
If you don’t feel like working through this trick open the example boxes. It’s really better if you do it yourself (the first time) though.
Example
Let’s choose $5701$.Let’s say the number has $n$ digits. We are going to make $n$ “layers” of numbers. The first layer only has one number in it: the number you chose.
Copy this number into the second layer too. The other number in the second layer is a “flipped” version of your chosen number. To do this, take off the most significant digit of your number and subtract the rest. So if you chose $2345$, calculate $2000  345 = 1655$. These two numbers form the second layer.
Example
We expect to have four layers. Our first two look like this:
Layer  Numbers  Notes 

1^{st}  $\left\{5701\right\}$  The chosen number 
2^{nd}  $\left\{5701,4299\right\}$  $5000701 = 4299$ 
Repeat this process to get the $(i+1)$^{th} layer from the $i$^{th}: first copy down all the numbers in the $i$^{th} layer into the $(i+1)$^{th}. Then add all the numbers “flipped” at the $i$^{th} digit. So to go from layer two to three with the number $2345$ we also add $2300  45$. If we were going from layer 3 to layer 4 we’d add $23405 =2335$.
One other thing: don’t do this if the $i$^{th} digit is 0. So you can’t flip the number $2034$ at the $2$^{nd} digit to get $2000  34$. Also, if you ever end up with duplicate values in a layer, just ignore them.
Example
We can fill in the rest of the layers:
Layer  Numbers  Notes 

1^{st}  $\left\{5701\right\}$  The chosen number 
2^{nd}  $\left\{5701,4299\right\}$  $5000701 = 4299$ 
3^{rd}  $\left\{5701,5699,4299,4101\right\}$  $57001 = 5699$ and $420099=4101$ 
4^{th}  $\left\{5701,5699,5681,4299,4281,4101\right\}$  We ignore $57001 = 5699$ and $41001=4099$ because we have a rule not to flip if the flipping digit is zero. 
Once this is done, you should end up with quite a few numbers in the $n$^{th} layer. For each of these numbers, perform the following operation:
 Subtract $1$
 Add one to each of the digits
 Multiply these together
So if the number $2340$ was in the last layer, we look at $2339$, add one to each of the digits to get $3$, $4$, $4$ and $10$ and multiply these together to get $480$.
Example
The last layer was $\left\{5701,5699,5681,4299,4281,4101\right\}$. Here is the result of the digit operation:
Number  Result of operation 

$5701$  $57011=5700$ so we add 1 to each digit and multiply to get $6\times 8\times 1 \times 1 = 48$ 
$5699$  $56991=5698$ so we calculate $6\times 7\times 10\times 9 = 3780$ 
$5681$  $56811=5680$ so we calculate $6\times 7\times 9\times 1 = 378$ 
$4299$  $42991=4298$ so we calculate $5\times 3\times 10\times 9 = 1350$ 
$4281$  $42811=4280$ so we calculate $5\times 3\times 9\times 1 = 135$ 
$4101$  $41011=4100$ so we calculate $5\times 2\times 1\times 1 = 10$ 
Finally, add all the numbers from the last step together…
Example
$$48+ 3780+378+1350+135+10=5701$$Is this your number?
If I’m honest. I don’t know why this works.
The game is reminiscent of the “pick a number, any number” or “shoe size trick”. In that incarnation, you are asked to think of any number (which we’ll call $x$), double it (to get $2x$), add sixteen (for $2x+16$), then divide by two (now $x+8$) before subtracting the number you originally thought of. Its moderately unsurprising that you end up with 8^{1}.
I, together with a coauthor, have a proof^{2} that this works if you write numbers base $p$, but not base 10 as we’ve done above. Even then its a very fiddly and complicated proof. An outline is as follows:
 We construct a particular vector space (or “module”) of dimension $n$ called $V_n$.
 We then write it as the direct sum of a bunch of other modules called $U_i$ where the $i$ are those numbers appearing in the last layer above: $V_n = \bigoplus_i U_i$.
 Then we show that the strange digit operation gives the dimension of $U_i$ (so $U_{230} = 3\times 4\times 1 = 12$).
 Since the direct sum simply sums the dimensions we have the result: $n = \dim V_n = \sum_i \dim U_i$.
However, this brushes a ton of representation theory under the carpet. I’d love to hear any explanation for why this works that doesn’t use high powered maths.

Incidentally, my friends and I once shortcircuited a escape room puzzle by doing the algebra instead of the arithmetic. The GM was rather confused… ↩︎