Most logic inputs will have a specified maximum current rating. This is the allowable maximum current that can be "forced" into the input without causing problems such as latch up (for instance). It's usually in the realm of 1 mA. To force current into an input means that the input voltage has to exceed the IC voltage supply by (usually) 0.3 volts.
So, if the voltage divider has a Thevenin resistance of 2k22 (6k8 || 3k3), in order to pass 1 mA the unloaded voltage from the potential divider has to be 3.6 volts plus 1mA*2k2 = 5.82 volts.
In effect that means that the input voltage to a 6k8 and 3k3 voltage divider is 17.81 volts. That's a lot more than 10 volts. Is that good enough? It usually is but, there's nothing wrong in placing a 10 kΩ resistor between the potential divider and the logic input for good measure.
A 10 k resistor would drop 10 volts when 1 mA is flowing and this vastly improves the over-voltage withstand capabilities of your circuit.
simulate this circuit – Schematic created using CircuitLab
If you were really OCD about it you could add a TVS diode across the 3k3 resistor to limit the voltage to something like 15 volts. That would be fairly bombproof and it's likely the 6k8 would burn before anything bad happened to your precious silicon!!