Four jewellers had respectively 8 rubies, 10 sapphires, 100 pearls and 5 diamonds.
Each gave one gem from their collection to each of the rest.
Afterwards they noticed that they all had collections of gems of precisely equal value.
Can you work out the relative values of each gem?
Can you then work out how much each jeweller gained or lost?
The first thing I tried was to set up a system of equations. If each jeweller has given away three gems and received one of the others, we know the following four quantities are the same:
$5r + s + p + d$
$r + 7s + p + d$
$r + s + 97p + d$
$r + s + p + 2d$
Looking at the first pair, the difference between them is $4r - 6s$ -- which has to be 0. Similarly, $6s = 96p$ and $96p = d$.
If we pick $p$ as our base unit, a diamond is worth $96p$, a sapphire a sixth of that, or $16p$, and a ruby works out to be $24p$, which gives the relative values of the gems.
Everybody gains $r + s + p + d - 4g$, where $g$ is the value of their own gem; this works out to $137p - 4g$. The ruby-cutter gains $41p$, the sapphire jeweller gains $73p$, the poor old diamond-seller loses $247p$ and the lucky pearl swine gains $133p$. A quick sanity check: $41 + 73 + 133 - 247 = 0$, so it's at least a plausible answer.
A neat short-cut to solving the whole thing is to note that if everybody throws away the new jewels they've received as well as one of their own, they will all have lost the same amounts, and their holdings will still be equal to each other. This is the same as each merchant just binning four jewels to begin with, which leads directly to the statement $4r = 6s = 96p = d$ with hardly any working out.
Philosophical aside: I reckon many people think of the first way as "proper maths" - look! Lots of equations! In reality, the second way is more what I think of as mathematical: a moment's thought simplifies the situation and turns it into something very straightforward. I'd characterise it as intelligent laziness; rather than solving the puzzle in front of you, it's finding an equivalent puzzle that's less work.
* Thanks to @ajk44 and Nrich maths for the puzzle.