In the expansion of $(2x +3y)^{20}$, find the number of coefficients divisible by $144$.

[i]Proposed by Hannah Shen[/i]
To solve the problem, we need to find the number of coefficients in the expansion of $(2x + 3y)^{20}$ that are divisible by $144$. 

1. **Express $144$ in terms of its prime factors:**
   \[
   144 = 2^4 \times 3^2
   \]
   This means that for a coefficient to be divisible by $144$, it must contain at least four factors of $2$ and two factors of $3$.

2. **Use the Binomial Theorem to expand $(2x + 3y)^{20}$:**
   The general term in the expansion of $(2x + 3y)^{20}$ is given by:
   \[
   T_{k} = \binom{20}{k} (2x)^{20-k} (3y)^k
   \]
   Simplifying the term:
   \[
   T_{k} = \binom{20}{k} 2^{20-k} 3^k x^{20-k} y^k
   \]
   We are interested in the coefficient:
   \[
   \text{Coefficient} = \binom{20}{k} 2^{20-k} 3^k
   \]

3. **Determine the conditions for divisibility by $144$:**
   - The coefficient $\binom{20}{k} 2^{20-k} 3^k$ must be divisible by $2^4 \times 3^2$.
   - This implies:
     \[
     2^{20-k} \geq 2^4 \quad \text{and} \quad 3^k \geq 3^2
     \]
   - Simplifying these inequalities:
     \[
     20 - k \geq 4 \implies k \leq 16
     \]
     \[
     k \geq 2
     \]

4. **Count the valid values of $k$:**
   - $k$ must satisfy $2 \leq k \leq 16$.
   - The number of integers $k$ in this range is:
     \[
     16 - 2 + 1 = 15
     \]

Conclusion:
The number of coefficients in the expansion of $(2x + 3y)^{20}$ that are divisible by $144$ is $\boxed{15}$.