Sizing the pulldown resistors for 74HCT08 logic inputs
If you know nothing about CMOS, you need ohms law, a data sheet, and some common sense.

If you know something about CMOS, then just the common sense will do.

Per the data sheet, the input leakage current for a 74HCT08 is \$1 \mathrm{\mu A}\$.  Per ohms law, this means we need \$R &lt; \mathrm{\frac{0.9V}{1\mu A} = 900k\Omega}\$.

Common sense says that unless we're designing something for insanely low current consumption, and for which we can guarantee an electrically quiet environment, using the maximum resistance here is silly.  So choose something convenient, like \$\mathrm{10k\Omega}\$.

A basic knowledge of CMOS says that for all practical purposes a CMOS chip's input current is nothing -- so you fall back on common sense, and you're done.

Note: that this is not the case for a lot of modern microcontrollers -- they are often designed so that without the pins being programmed, they have built-in weak pull-ups, to reduce current, and for historical reasons (see my final note, below).  But sometimes they have built-in weak pull-downs on some pins, for practical reasons.  So always read the data sheet.

Final note:  If you were designing circuits 30 years ago, then the gate in question may well have been TTL.  In that case, you would want a pull-up resistor (because TTL inputs source a bit of current, but sink less).  Then you'd use the same \$\mathrm{10k\Omega}\$, but it would be to VCC.  There is no reason not to do this for CMOS, and it's oh-so-slightly more comfortable for old circuit designers when they see it.