ABC is a plane lamina in the shape of an equilateral triangle. D and E are the midpoints of AB and AC, and G is the centroid of the lamina. The moment of inertia of the lamina about an axis passing through G and perpendicular to the plane ABC is I₀. If the part ADE is removed, the moment of inertia of the remaining part about the same axis is (NI₀)/(16), where N is an integer. The value of N is ______.

Let the triangle have mass M and side a. For an equilateral lamina about the perpendicular axis through its centroid, I₀=(Ma²)/(12).

ADE is the small triangle at the apex, similar to ABC with every length halved, so it has side a/2 and, since area goes as the square of the length, mass M/4.

About its own centroid: ((M/4)(a/2)²)/(12)=(Ma²)/(192)=(I₀)/(16).

But the axis we need runs through G, not through the centroid of ADE. Place B at the origin, C at (a,0) and A at (a/2,(a√3)/2).

Then G=(a/2,(a√3)/6), and the centroid of ADE works out to (a/2,(a√3)/3).

They lie on the same vertical line, a distance d=(a√3)/3-(a√3)/6=(a√3)/6 apart, so d²=(a²)/(12).

ADE about the axis through G: (Ma²)/(192)+M/4·(a²)/(12)=(Ma²)/(192)+(4Ma²)/(192)=(5Ma²)/(192).

Since I₀=(Ma²)/(12)=(16Ma²)/(192), that removed piece is 5/(16)I₀.

Remaining: I₀-5/(16)I₀=(11)/(16)I₀.

N=11.