Let $m$ and $n$ be arbitrary positive integers, and 
 $a, b, c$ be different natural numbers of the form $2^m.5^n$.  Determine
 the number of all equations of the form $ax^2-2bx+c=0$ if it is known that each equation
 has only one real solution.
To determine the number of equations of the form \( ax^2 - 2bx + c = 0 \) that have only one real solution, we need to analyze the discriminant of the quadratic equation. A quadratic equation \( ax^2 + bx + c = 0 \) has exactly one real solution if and only if its discriminant is zero.

1. **Discriminant Condition**:
   The discriminant \(\Delta\) of the quadratic equation \( ax^2 - 2bx + c = 0 \) is given by:
   \[
   \Delta = (-2b)^2 - 4ac = 4b^2 - 4ac
   \]
   For the equation to have exactly one real solution, the discriminant must be zero:
   \[
   4b^2 - 4ac = 0 \implies b^2 = ac
   \]

2. **Form of \(a\), \(b\), and \(c\)**:
   Given that \(a\), \(b\), and \(c\) are of the form \(2^m \cdot 5^n\), we can write:
   \[
   a = 2^m \cdot 5^n, \quad b = 2^i \cdot 5^j, \quad c = 2^k \cdot 5^l
   \]
   where \(m, n, i, j, k, l\) are non-negative integers.

3. **Condition \(b^2 = ac\)**:
   Substituting the forms of \(a\), \(b\), and \(c\) into the equation \(b^2 = ac\), we get:
   \[
   (2^i \cdot 5^j)^2 = (2^m \cdot 5^n)(2^k \cdot 5^l)
   \]
   Simplifying, we obtain:
   \[
   2^{2i} \cdot 5^{2j} = 2^{m+k} \cdot 5^{n+l}
   \]
   For the equality to hold, the exponents of 2 and 5 on both sides must be equal:
   \[
   2i = m + k \quad \text{and} \quad 2j = n + l
   \]

4. **Parity Conditions**:
   From the equations \(2i = m + k\) and \(2j = n + l\), we see that \(m\) and \(k\) must have the same parity (both even or both odd), and similarly, \(n\) and \(l\) must have the same parity.

5. **Infinite Solutions**:
   Since \(m\), \(n\), \(i\), \(j\), \(k\), and \(l\) are arbitrary non-negative integers, there are infinitely many combinations that satisfy the conditions \(2i = m + k\) and \(2j = n + l\). For any valid solution \((a, b, c)\), we can generate new solutions by multiplying \(a\), \(b\), and \(c\) by any power of 2 or 5, leading to infinitely many solutions.

The final answer is \( \boxed{ \text{There are infinitely many equations of the form } ax^2 - 2bx + c = 0 \text{ with exactly one real solution.} } \)