$4 \cdot 21$ Find the sum of all divisors $d=2^{a} \cdot 3^{b}(a, b$ being natural numbers $)$ of $N=19^{88}-1$.
(Austrian Mathematical Competition, 1988)
[Solution]
$$
\begin{aligned}
N= & 19^{88}-1 \\
= & (20-1)^{88}-1 \\
= & (4 \cdot 5-1)^{88}-1 \\
= & -C_{88}^{1} 4 \cdot 5+C_{88}^{2} 4^{2} \cdot 5^{2}-C_{88}^{3} 4^{3} \cdot 5^{3}+\cdots \\
& -C_{88}^{87} 4^{87} 5^{87}+C_{88}^{88} 4^{88} 5^{88} \\
= & -2^{5} \cdot 55+2^{6} M \\
= & 2^{5}(-55+2 M)
\end{aligned}
$$

where \( M \) is an integer.
Therefore, in the standard factorization of \( N \), the highest power of 2 is 5.
On the other hand,
$$
\begin{aligned}
N & =(2 \cdot 9+1)^{88}-1 \\
& =C_{88}^{1} 2 \cdot 9+C_{88}^{2} 8^{2} \cdot 9^{2}+\cdots+C_{88}^{88} 2^{88} \cdot 9^{88} \\
& =3^{2} \cdot 2 \cdot 88+3^{3} \cdot T \\
& =3^{2}(2 \cdot 88+3 T)
\end{aligned}
$$

where \( T \) is an integer.
Therefore, in the standard factorization of \( N \), the highest power of 3 is 2. That is,
$$
N=2^{5} \cdot 3^{2} \cdot L \text {. }
$$

where \( L \) does not contain the factors 2 and 3.
Thus, the sum of all divisors of \( N \) of the form \( d=2^{a} \cdot 3^{b} \) is
$$
\left(2+2^{2}+2^{3}+2^{4}+2^{5}\right)\left(3+3^{2}\right)=744 .
$$