How can be $\frac{1}{4\pi R^2}\int_{S}V_{ext}(R)da= V_{ext}(0)$ physically explained? I was working out problem 3.1 (4th edition) of Introduction to Electrodynamics by Griffiths, which asks you for the average potential over a spherical surface due to a charge located inside the sphere (as well as verifying an equation). 
I understand that, mathematically, one gets for $V_{av}$:
$$V_{av} = \frac{q}{4 \pi \epsilon R}$$
And if there is a bunch of $q$ inside the sphere one gets:
$$V_{av} = \frac{Q_{enc}}{4 \pi \epsilon R}$$
Griffiths shows that the average potential due to exterior charges is the same as if they were placed at the center.
I understand the Math to get such a result but I do not understand this result physically speaking. 
How can be the average potential due to an external (out of the sphere) charge $q$ be equal to the average potential due to a charge $q$ located at the center?
I agree that the potential does not have a physical meaning as such, but my intuition tells me that the difference between position a (say its value at the center of the sphere) and b (say its value at the position out of the sphere) should matter. Besides, the potential falls off like $1/r$. Thus I do not see how is it possible (physically speaking) that:
$$V_{ext}(R)= V_{ext}(0)$$
 A: 
How can be the average potential due to an external (out of the sphere) charge $q$
  be equal to the average potential due to a charge $q$ located at the center?

It's not.  The average potential over a spherical surface of radius $R$ due to a point charge a distance $z > R$ from the center of the sphere is $$V_\text{ave} = \frac{q}{4 \pi \epsilon_0 z}
$$
(see §3.1.4 of Griffiths).  Meanwhile, the average potential over the surface if $z < R$ (i.e., the charge is inside the surface) is
$$
V_\text{ave} = \frac{q}{4 \pi \epsilon_0 R}.
$$
These quantities are not the same.
A: 1) I think you mean that "the average potential due to exterior charges is the same as the potential at the center". We're not talking about a different charge configuration, we're talking about some aspect about the potential.
2) $V_{ext}(R)= V_{ext}(0)$ is not the equation you're looking for, and is why your intuition is failing you. We're not saying that "the potential at $R$ is the same as the potential at $0$, we're saying that the average potential over $R$ is equal to the potential at $0$.
$$\frac{1}{4\pi R^2}\int_{S}V_{ext}(R)da= V_{ext}(0) \tag{1}$$
3) Edits. The term "average" is really what's important here. Taking an example of the $1/r$ potential, the contribution of $V$ is "balanced" around the sphere in such a way that $(1)$ is true. If $r$ is smaller, then $V$ is large, but it contributes to less surface area on the sphere, and similarly, if $r$ is large and $V$ is small, then the contribution is large. 
On the other hand, Laplace's equation intuitively tells us that the potential will be "smooth", with no local extrema anywhere. I like to imagine that the potential tried to be as boring as possible, hence the behavior over a surface will reflect the behavior of the inside; the field doesn't "fluctuate". It's not surprising that the average of that behavior tells us something about the value of the field within. 
