How many positive integers a less than $100$ such that $4a^2 + 3a + 5$ is divisible by $6$.
1. Define the function \( f(a) = 4a^2 + 3a + 5 \).
2. We need to find the values of \( a \) such that \( f(a) \) is divisible by \( 6 \). This means \( f(a) \equiv 0 \pmod{6} \).
3. We will check the values of \( f(a) \) for \( a \) modulo \( 6 \).

   - If \( a \equiv 0 \pmod{6} \):
     \[
     f(a) = 4(0)^2 + 3(0) + 5 \equiv 5 \pmod{6}
     \]
     Thus, \( f(a) \not\equiv 0 \pmod{6} \).

   - If \( a \equiv 1 \pmod{6} \):
     \[
     f(a) = 4(1)^2 + 3(1) + 5 = 4 + 3 + 5 = 12 \equiv 0 \pmod{6}
     \]
     Thus, \( f(a) \equiv 0 \pmod{6} \).

   - If \( a \equiv 2 \pmod{6} \):
     \[
     f(a) = 4(2)^2 + 3(2) + 5 = 16 + 6 + 5 = 27 \equiv 3 \pmod{6}
     \]
     Thus, \( f(a) \not\equiv 0 \pmod{6} \).

   - If \( a \equiv 3 \pmod{6} \):
     \[
     f(a) = 4(3)^2 + 3(3) + 5 = 36 + 9 + 5 = 50 \equiv 2 \pmod{6}
     \]
     Thus, \( f(a) \not\equiv 0 \pmod{6} \).

   - If \( a \equiv 4 \pmod{6} \):
     \[
     f(a) = 4(4)^2 + 3(4) + 5 = 64 + 12 + 5 = 81 \equiv 3 \pmod{6}
     \]
     Thus, \( f(a) \not\equiv 0 \pmod{6} \).

   - If \( a \equiv 5 \pmod{6} \):
     \[
     f(a) = 4(5)^2 + 3(5) + 5 = 100 + 15 + 5 = 120 \equiv 0 \pmod{6}
     \]
     Thus, \( f(a) \equiv 0 \pmod{6} \).

4. From the above calculations, \( f(a) \equiv 0 \pmod{6} \) when \( a \equiv 1 \pmod{6} \) or \( a \equiv 5 \pmod{6} \).

5. We need to count the number of positive integers \( a \) less than \( 100 \) that satisfy \( a \equiv 1 \pmod{6} \) or \( a \equiv 5 \pmod{6} \).

   - For \( a \equiv 1 \pmod{6} \):
     The sequence is \( 1, 7, 13, \ldots, 97 \). This is an arithmetic sequence with the first term \( a_1 = 1 \) and common difference \( d = 6 \).
     The \( n \)-th term of the sequence is given by:
     \[
     a_n = 1 + (n-1) \cdot 6 = 6n - 5
     \]
     Setting \( 6n - 5 < 100 \):
     \[
     6n < 105 \implies n < 17.5
     \]
     Thus, \( n \) can be \( 1, 2, \ldots, 16 \), giving us \( 16 \) terms.

   - For \( a \equiv 5 \pmod{6} \):
     The sequence is \( 5, 11, 17, \ldots, 95 \). This is an arithmetic sequence with the first term \( a_1 = 5 \) and common difference \( d = 6 \).
     The \( n \)-th term of the sequence is given by:
     \[
     a_n = 5 + (n-1) \cdot 6 = 6n - 1
     \]
     Setting \( 6n - 1 < 100 \):
     \[
     6n < 101 \implies n < 16.83
     \]
     Thus, \( n \) can be \( 1, 2, \ldots, 16 \), giving us \( 16 \) terms.

6. Adding the number of terms from both sequences, we get:
   \[
   16 + 16 = 32
   \]

The final answer is \(\boxed{32}\).