Quantum chromodynamics (the theory which describes the strong interaction) has a feature called colour confinement, which says that quarks will favour being in colourless configuration (where colourless = zero colour charge). The two easiest ways for this to happen are in mesons (two quarks: one quark and one antiquark of the same colour, like red + antired = colourless), or in baryons (three quarks: 1 red + 1 green + 1 blue = colourless) like the familiar proton and neutron. However other configurations are possible, these are just the simplest ones. In certain incredibly difficult to attain conditions, we can avoid producing either a two- or three-quark composite particle but produce instead a five-quark particle (e.g. 1 red + 1 green + 1 blue + 1 red + 1 antired = colourless).

This "colour" means colour charge, it doesn't have any relation to the regular meaning of "colour of light".

> Oh, so quarks can only be composites of two, three, and five?

They can also be in fours. Possibly other configurations; the four and five quark configurations were theorized in 1964 but only confirmed in 2014 & 2015.

> Any reason that could be stated in layman's terms?

None at all, in fact, the source article does so: “In the conventional quark model, composite particles can be either mesons formed of quark–antiquark pairs or baryons formed of three quarks. Particles not classified within this scheme are known as exotic hadrons. When Murray Gell-Mann proposed the quark model in his fundamental 1964 paper, he mentioned the possibility of exotic hadrons such as pentaquarks, but it took 50 years to demonstrate their existence experimentally.”

(There is slightly more detail on the read more link, too.)

Basically, take the numbers 1/3, -1/3, and count how many terms you need to produce something that sums to a whole number. You can get four with (1/3 + 1/3 - 1/3 - 1/3), so it's not an impossible configuration.

If you're willing to dive in a bit, Matt Strassler has a nice blog aimed at the lay reader. His mini-series on the structure of the proton is quite excellent:

https://profmattstrassler.com/articles-and-posts/largehadron...

> You can get five with (1/3 + 1/3 + 1/3 + 1/3 - 1/3).

I'm sorry; although you very clearly referred to winding up with a whole number, I somehow read it as winding up with 0.