Triangles in Triangles in Triangles...

The answer is $5$ . Here is a non-rigorous and lazy solution. Enjoy!

We'll show you can make subdivide our equilateral triangle into $6, 7$ and $8$ not necessarily congruent equilateral triangles and then show that you cannot subdivide it into $5$ .

Part 1: Subdividing Equilateral Triangles into $6 , 7$ and $8$ necessarily congruent equilateral triangles .

Suppose we have an equilateral triangle with each side of length $a$ . Here's how to subdivide it into $6, 7$ and $8$ respectively :

Disclaimer : I didn't have a ruler or compass on me when drawing these images (what am I doing with my life?)!

Part 2: Showing that If I can subdivide into $K$ triangles I can subdivide into $K+3$ .

Suppose we can subdivide an equilateral triangle into $K$ not necessarily congruent equilateral triangles. Then we can also subdivide an equilateral triangle into $K + 3$ not necessarily congruent equilateral triangles by making one of our subdivided triangles into $4$ like so :

Using the cases $K = 6,7,8$ from before and this information we can prove by induction on $K$ that we can make every $K \equiv 0 \mod(3),K \equiv 1 \mod(3),K \equiv 2 \mod(3)$ for all $K \ge 6$ .

Part 3: Finishing up.

Now we've shown we can subdivide for all for all $K \ge 6$ it suffices to show that we can NOT do so for $K = 5$ . I'm not going to show that here (at least not yet! ) but hopefully you can prove for yourselves that this is not possible. If you can post it for everyone to see!