A-Level Mechanics Ladder Friction Question. Is the question wrong? I received this question in one of my math homeworks. To sum the past couple months into a short sentence, this question has caused a lot of arguments:
A uniform ladder, of weight 'W' and length 5m, has one end on rough horizontal ground and the other touching a smooth vertical wall. The coefficient of friction between the ladder and the ground is 0.3 .
The top of the ladder touches the wall at a point 4m vertically above the level of the ground.
A) Show that the ladder can not rest in equilibrium in this position.
In order to enable the ladder to rest in equilibrium in the position described above, a brick is attached to the bottom of the ladder.
Assuming that this brick is at the lowest point of the ladder, but not touching the ground:
B) Show that the horizontal frictional force exerted by the ladder on the ground is independent of the mass of the brick.
C) Find, in terms of W and g, the smallest mass of the brick for which the ladder will rest in equilibrium.
Now, that's everything the question says, I left it all in so that no details are left out. The problem I believe, lies in part 'B' and 'C'. Without even thinking about it mathematically yet, do they not contradict each other? To start with, it mentions to put the ladder into equilibrium, the brick is added. But then it says, in part B, to show that the brick has no effect on the frictional force (but as far as I am aware this is the only force keeping it from slipping). But finally, it says in part C to show the minimum mass of the brick to keep the ladder in equilibrium. Surely, if the mass of the brick is keeping the ladder in equilibrium, the mass of the brick can not be independent of the frictional force.
To get into the maths:


*

*The ladder slips down the wall due to the weight of the ladder. Since the wall is smooth, it cannot exert a vertical force to oppose this. So, the only force that can stop it slipping is the frictional force exerted on the floor.

*The frictional force is: F=μR. Before the brick is on the ladder, this will be: F= 0.3*W , as all the weight of the ladder will be taken on by the reaction force between the ground and the bottom of the ladder.

*When the brick is added, Friction will now be: F= 0.3 * (W + (Mass of brick * g)).


So does this not mean that the frictional force is dependent on the mass of the brick? If it isn't, what puts the ladder into equilibrium when the brick is added?
If you can see anything that i'm missing, that would be great, this has been bugging me for far too long. I have a further maths teacher and an entire class who still disagrees with me. Managed to convince my maths teacher, whilst my physics teacher just laughed when he saw the question.
Hopefully this all makes sense, writing it all down is a lot harder than explaining in person.
Thanks.
 A: What they mean in B is that the horizontal force necessary to stop the ladder from sliding is the same with or without the brick.   If the ladder without the brick were capable of exerting this force horizontally, it would not slip.  With the added downforce of the brick, that amount of horizontal force is available, so the ladder does not slip.  
The coefficient of friction really sets a maximum amount of friction available to stop a slip or the resistive force if a slip is occurring.  If the ladder were vertical, the coefficient of friction would still be $0.3$, but the sideways force would be smaller than $0.3W$ (note there is no $g$ if you are given the weight instead of the mass) and it would not slip.
A: Let $m$ be the mass of the brick
So $R=W+mg$
and the frictional force is 
$$F= 0.3(W+mg)x $$
Where $0 \leq x \leq 1$
Let $\theta$ be  the angle between the ladder and the ground, so $\sin \theta = 0.8$ and $ \cos \theta =0.6$
Resolving torques about the point of contact with the wall, there are three torques turning the ladder in the direction of the wall ...
$$ \tau_1= 2.5W\cos\theta + 5 mg \cos \theta +  5F \sin\theta \\=1.5W + 3mg+1.2x(W+mg)$$
And one torque turning in the direction away from the wall coming from the reaction force.
$$ \tau_2= 5(W+mg) \cos \theta =4(W+mg)$$
Equilibrium imposes the following solution on $x$ ...
$$ x = 1.25 \frac W{W+mg}$$Since $x\le 1$ we have....
$$mg\ge 0.25W$$
So not possible for $m=0$ 
Whenever equilibrium is achieved
$$F= 0.3(W+mg)x =0.375W$$
independent of the mass of the brick.
