jmc

It's clear that we must have $a=2^j{5}^k$, $b=2^m{5}^n$ and $c=2^p{5}^q$ for some nonnegative integers $j,k,m,n,p,q$. Dealing first with the powers of 2: from the given conditions, $\max (j,m)=3$, $\max (m,p)=\max (p,j)=4$. Thus we must have $p=4$ and at least one of $m,j$ equal to 3. This gives 7 possible triples $(j,m,p)$: $(0,3,4),(1,3,4),(2,3,4),(3,3,4),(3,2,4),(3,1,4)$ and $(3,0,4)$. Now, for the powers of 5: we have $\max (k,n)=\max (n,q)=\max (q,k)=3$. Thus, at least two of $k,n,q$ must be equal to 3, and the other can take any value between 0 and 3. This gives us a total of 10 possible triples: $(3,3,3)$ and three possibilities of each of the forms $(3,3,n)$, $(3,n,3)$ and $(n,3,3)$. Since the exponents of 2 and 5 must satisfy these conditions independently, we have a total of $7\cdot 10=\boxed{70}$ possible valid triples.