13. (19th Austrian Mathematical Competition) Find $N=$ $19^{88}-1$ for all divisors $d=2^{2} \cdot 3^{b}$ (where $a, b$ are natural numbers) and determine the sum of these divisors $d$.
$$
\begin{aligned}
N & =(20-1)^{88}-1 \\
& =(1-4 \times 5)^{88}-1 \\
& =-C_{88}^{1} 4 \times 5+C_{88}^{2} 4^{2} \times 5^{2} \\
& -C_{88}^{8} 4^{8} \times 5^{3}+\cdots \\
& -C_{88}^{87} 4^{87} \times 5^{87}+C_{88}^{88} 4^{88} \times 5^{88} \\
& =-2^{5} \times 55+2^{8} M \\
& =2^{5}(-55+2 M),
\end{aligned}
$$

where $M$ is an integer.
Therefore, the highest power of 2 in the prime factors of $N$ is 5.
On the other hand,
$$
\begin{aligned}
N & =(1+2 \times 9)^{88}-1 \\
& =C_{88}^{1} 2 \times 9+C_{88}^{2} 2^{2} \times 9^{2}+\cdots \\
& +C_{88}^{88} 2^{88} \times 9^{88} \\
& =3^{2} \times 2 \times 88+3^{3} T \\
& =3^{2}(2 \times 88+3 T),
\end{aligned}
$$

where $T$ is an integer.
Therefore, the highest power of 3 in the prime factors of $N$ is 2.
Thus, $N=2^{5} \cdot 3^{2} \cdot L$, where $L$ does not contain 2 and 3 as factors.

Therefore, the sum of all divisors of $N$ of the form $d=2^{\mathrm{a}} \cdot 3^{\mathrm{b}}$ is
$$
\left(2+2^{2}+2^{3}+2^{4}+2^{5}\right)\left(3+3^{2}\right)=744 .
$$