Positive integers $a$, $b$, and $c$ are all powers of $k$ for some positive integer $k$. It is known that the equation $ax^2-bx+c=0$ has exactly one real solution $r$, and this value $r$ is less than $100$. Compute the maximum possible value of $r$.
**
   For a quadratic equation \(ax^2 + bx + c = 0\) to have exactly one real solution, the discriminant must be zero. The discriminant \(\Delta\) of the quadratic equation \(ax^2 - bx + c = 0\) is given by:
   \[
   \Delta = b^2 - 4ac
   \]
   Since the equation has exactly one real solution, we have:
   \[
   b^2 - 4ac = 0 \implies b^2 = 4ac
   \]

3. **Expressing \(a\), \(b\), and \(c\) as Powers of \(k\):**
   Let \(a = k^m\), \(b = k^n\), and \(c = k^\ell\) for some integers \(m\), \(n\), and \(\ell\). Substituting these into the discriminant condition, we get:
   \[
   (k^n)^2 = 4(k^m)(k^\ell) \implies k^{2n} = 4k^{m+\ell} \implies k^{2n} = k^{m+\ell + 2}
   \]
   Since the bases are the same, we can equate the exponents:
   \[
   2n = m + \ell + 2
   \]

4. **Finding the Value of \(r\):**
   The solution \(r\) of the quadratic equation \(ax^2 - bx + c = 0\) is given by:
   \[
   r = \frac{b}{2a}
   \]
   Substituting \(a = k^m\) and \(b = k^n\), we get:
   \[
   r = \frac{k^n}{2k^m} = \frac{k^n}{k^{m+1}} = k^{n - m - 1}
   \]
   We need to maximize \(r = k^{n - m - 1}\) such that \(r < 100\).

5. **Choosing \(k\) and Solving for Maximum \(r\):**
   To maximize \(r\), we choose \(k = 2\) (since powers of 2 grow quickly and are easy to handle). We need to find the largest power of 2 less than 100:
   \[
   2^6 = 64 \quad \text{(since } 2^7 = 128 \text{ is greater than 100)}
   \]
   Therefore, we set:
   \[
   r = 2^{n - m - 1} = 64 \implies n - m - 1 = 6 \implies n - m = 7
   \]

6. **Constructing \(a\), \(b\), and \(c\):**
   We also need to satisfy the equation \(2n = m + \ell + 2\). Using \(n - m = 7\), we can express \(n\) as \(n = m + 7\). Substituting this into the equation:
   \[
   2(m + 7) = m + \ell + 2 \implies 2m + 14 = m + \ell + 2 \implies m + 12 = \ell \implies \ell = m + 12
   \]
   We can choose \(m = 1\) (for simplicity), then:
   \[
   n = 1 + 7 = 8 \quad \text{and} \quad \ell = 1 + 12 = 13
   \]
   Thus, we have:
   \[
   a = 2^1 = 2, \quad b = 2^8 = 256, \quad c = 2^{13} = 8192
   \]

7. **Verification:**
   The quadratic equation becomes:
   \[
   2x^2 - 256x + 8192 = 0
   \]
   Solving for \(x\):
   \[
   x = \frac{256}{2 \cdot 2} = \frac{256}{4} = 64
   \]
   This confirms that \(r = 64\) is a solution.

The final answer is \(\boxed{64}\)