So my question is regarding where the Fermi energy is when you have 2D electron gas in an applied magnetic field. My book explains that, using the Landau gauge, you find that the 2D density of states (normally the constant $g_{2D} = m/\pi \hbar^2$) collapses into a series of Dirac delta functions $g_{2D}\delta(E - E_n)$ for the harmonic oscillator levels $E_n = \hbar \omega_c (n + 1/2)$ ($\omega_c$ being the cyclotron frequency) because in this gauge, all the states with the same $n$ are degenerate. This is illustrated in the top half of this picture.
They also say that, due to scattering, the peaks can't be defined with such high precision, so you actually get shapes more like the bottom half of that picture. This all makes sense to me.
But I get confused when I try to find the Fermi energy for a given density and nonzero magnetic field (at $T = 0$, always...) without the scattering that widens the peaks. With no magnetic field, I would simply integrate the 2D DoS from $0$ to $E_F$ and set that equal to my number density, and solving for $E_F$, I'd find it somewhere. But with a magnetic field I run into a problem: My understanding of the Fermi energy is that it's the energy such that all states at or below it are occupied and all those above are unoccupied.
But this seems to only offer certain values of the number density for this case! When you integrate over a range of these Dirac deltas, you essentially pick up $g_{2D}\hbar \omega_c$ states for each delta you integrate over, so it seems like in the case of an applied B field, $T = 0$, and no scattering that widens the deltas to something more realistic, you can only have number densities of the form $n_{2D} = m g_{2D}\hbar \omega_c$, where $m$ is an integer.
I'm sure I'm not explaining this well and that I'm probably wrong, but what am I missing?