How many numbers satisfy this property?

Notice that $2^n \cdot 5^n = 10^n$ .
Let's write $2^n$ and $5^n$ in scientific notation.
Now, $2^n = c_1 \cdot 10^{k_1}, 5^n = c_2 \cdot 10^{k_2}$ , where $1 \leq c_1, c_2 < 10$ .
Since $2^n$ and $5^n$ start with the same digit, $c_1$ and $c_2$ must have the same integer part, and thus $c_1 = x + f_1, c_2 = x + f_2$ , where $x$ is an integer, $f_1$ is the fractional part of $c_1$ and $f_2$ is the fractional part of $c_2$ .
Since $2^n \cdot 5^n = 10^n \implies c_1 \cdot c_2 = \left(x + f_1\right)\cdot\left(x + f_2\right) = x^2 + \left(f_1+f_2\right)x + f_1f_2 = 10$ , we know that $x = 3$ is the only solution because $0 < f_1, f_2 < 1$ .
Hence, both $2^n$ and $5^n$ must start with $3$ .

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First $n$ which satisfies the property is $5$ .

We get the next candidate by adding $10$ to the previous $n$ .
That is the next possible $n$ , since when we look at powers of $2$ , $2^{10} = 1024$ is the lowest power of $2$ which written in scientific notation has the coefficient between $1$ and $\dfrac43$ . (This is because we want the next candidate to start with $3$ )

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Using the above we see that $15$ satisfies this property, but $25$ doesn't.

Now we repeatedly add $10$ to the $n$ (exponent) until we reach a power of $2$ which when written in scientific notation has the coefficient over $\dfrac{30}8$ (This is because we want to lower the candidate little bit since repeatedly multiplying by $1024$ increased the value too much.), since $8 = 2^3$ has the coefficient closest to $10$ of all powers of $2$ below $1024$ .

When we reach such $n$ , we add $3$ to it and we have our next candidate.

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Using this approach, we get that following positive integers $n \leq 200$ satisfy the property: $5,15,78,88,98,108,118,181,191$ .

Hence, the answer is $\boxed{9}$ .