14.3.27 ** Find the sum of all divisors $d=2^{a} \cdot 3^{b}(a, b>0)$ of $N=19^{88}-1$.
Notice that $19^{88}-1=\left(19^{11}-1\right)\left(19^{11}+1\right)\left(19^{22}+1\right)\left(19^{44}+1\right)$,
(1) We will find the largest integers $m, n$ such that $2^{m} 3^{n} \mid 19^{88}-1$. Since $19^{11}-1=\left(2 \times 3^{2}+1\right)^{11}-1 \equiv$ $11 \times 2 \times 3^{2}(\bmod 27)$, it follows that $3^{2} \mid 19^{11}-1$, but $3^{3} \nmid 19^{11}-1$, and the other three factors on the right side of (1) $\equiv$ $2(\operatorname{mod} 3)$. Thus, $n=2$. Also, since $19^{11}+1=(4 \times 5-1)^{11}+1 \equiv 11 \times 4 \times 5(\bmod 8)$, $19^{44}+1 \equiv 19^{22}+1 \equiv 2(\bmod 4)$, it follows that $4\left|19^{11}+1,2\right| 19^{11}-1,2\left|19^{22}+1,2\right|$ $19^{44}+1$, but $8 \nmid 19^{11}+1,4 \nmid 19^{11}-1,4 \nmid 19^{22}+1,4 \nmid 19^{44}+1$. Thus, $m=5$. The required sum is $\left(3+3^{2}\right)\left(2+2^{2}+2^{3}+2^{4}+2^{5}\right)=744$.