Internal resistance of a digital multimeter

[obsoletepower]

Homework Statement

Provided that the input resistance of the DMM is
much larger than the resistance values in the circuit to
which it is connected, the DMM will not draw a
significant current. Estimate the input resistance of the
DMM by connecting the 10 M resistor supplied in
series with it and noting the voltage reading when
this series combination is connected to your DC black
box. Explain clearly the logic of doing this.

Basically I have a simple circuit with a DC black box which is a box with a 1.5V battery and three resistors (1 in series and 2 in parallel) and I have a digital multimeter that I have to estimate its internal resistance by connecting a 10 MΩ resistor to the circuit in series with the multimeter. I can't really think of why connecting such a large resistor would help determine the internal resistance.

The Attempt at a Solution

I haven't been able to come up with an answer because first I need to understand the role of the 10 MΩ resistor.

 

[mgb_phys]

Draw the meter as a meter + a 1M resistor in series.
Now connect a 10M resistor across the meter inputs.

What do you know about resitances in parallel

 

[obsoletepower]

Draw the meter as a meter + a 1M resistor in series.
Now connect a 10M resistor across the meter inputs.

What do you know about resitances in parallel

First of all, I appreciate your post.

Well, for resistances in parallel we know that the overall resistance will be smaller than any individual resistance. so in this case 1 / R = 1 / R_s + 1 / R_p where R_s is the series resistor and R_p is the parallel resistor. so R = 0.9M, roughly. I still don't get why the 10M resistor is needed.