Find the total number of primes $p<100$ such that $\lfloor (2+\sqrt{5})^p \rfloor-2^{p+1}$ is divisible by $p$. Here $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. 
1. **Identify the problem and given expression:**
   We need to find the total number of primes \( p < 100 \) such that \( \lfloor (2+\sqrt{5})^p \rfloor - 2^{p+1} \) is divisible by \( p \). Here, \( \lfloor x \rfloor \) denotes the greatest integer less than or equal to \( x \).

2. **Simplify the expression:**
   Consider the expression \( (2+\sqrt{5})^p + (2-\sqrt{5})^p \). Since \( 2-\sqrt{5} \) is a number between 0 and 1, \( (2-\sqrt{5})^p \) becomes very small as \( p \) increases. Therefore, \( \lfloor (2+\sqrt{5})^p \rfloor \) is approximately \( (2+\sqrt{5})^p + (2-\sqrt{5})^p \) minus a small fraction.

3. **Rewrite the expression:**
   \[
   \lfloor (2+\sqrt{5})^p \rfloor = (2+\sqrt{5})^p + (2-\sqrt{5})^p - \epsilon
   \]
   where \( 0 < \epsilon < 1 \).

4. **Subtract \( 2^{p+1} \) from the expression:**
   \[
   \lfloor (2+\sqrt{5})^p \rfloor - 2^{p+1} = (2+\sqrt{5})^p + (2-\sqrt{5})^p - 2^{p+1} - \epsilon
   \]

5. **Simplify the expression modulo \( p \):**
   Since \( 2-\sqrt{5} \) is very small, \( (2-\sqrt{5})^p \) is negligible modulo \( p \). Therefore, we can approximate:
   \[
   (2+\sqrt{5})^p + (2-\sqrt{5})^p \approx (2+\sqrt{5})^p \pmod{p}
   \]

6. **Use the binomial theorem:**
   \[
   (2+\sqrt{5})^p = \sum_{k=0}^{p} \binom{p}{k} 2^{p-k} (\sqrt{5})^k
   \]
   Since \( p \) is a prime, by Fermat's Little Theorem, \( 2^p \equiv 2 \pmod{p} \) and \( (\sqrt{5})^p \equiv (\sqrt{5}) \pmod{p} \).

7. **Combine the terms:**
   \[
   (2+\sqrt{5})^p + (2-\sqrt{5})^p = 2 \sum_{i=0}^{\frac{p-1}{2}} \binom{p}{2i} 2^{p-2i} 5^i
   \]
   This simplifies to:
   \[
   2 \cdot 2^p - 2^{p+1} \equiv 0 \pmod{p}
   \]

8. **Check the divisibility:**
   Since \( 2 \cdot 2^p - 2^{p+1} \equiv 0 \pmod{p} \), it follows that \( p \) divides the expression for all primes \( p \geq 3 \).

9. **Count the primes:**
   The primes less than 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Excluding 2, there are 24 primes.

The final answer is \(\boxed{24}\)