A question of Number Theory+Binomial Theorem Let the sum of all divisors of the form $2^{p} \cdot 3^{q}$ (p,q are positive integers) of the number $19^{88} - 1$ be $\lambda$ . Find unit digit of $\lambda$
I noticed one fact that $2^{p} \cdot 3^{q}$ means any number which is not of form $6k\pm 1$.
Tried to expand it $(19^{44}+1)(19^{22}+1)(19^{11}+1)(19^{11}-1)$ then found that $19^{11}+1$ has one factor as $20$ but this approach didn't work.
Then I tried to expand it using Binomial Theorem but it turned out be very lengthy process.

We just need to find the largest factor of $19^{88}-1$ of the form $2^p3^q$. The rest shall follow from considering its factors.
Using the factorization $19^{88}-1 = (19^{44}+1)(19^{22}+1)(19^{11}+1)(19^{11}-1)$, we first check the value of $p$. Taking mod $4$:
$$19^{88}+1 \equiv (-1)^{88}+1 \equiv 2 \equiv 19^{44}-1 \pmod 4$$
$$19^{11}+1 \equiv (-1)^{11}+1 \equiv 0 \pmod 4$$
$$19^{11}-1 \equiv (-1)^{11}-1 \equiv 2 \pmod 4$$
Now we check mod $8$ for $19^{11}+1$: $19^{11}+1 \equiv 361^5 \times 19 + 1\equiv 1^5 \times 3 + 1 \equiv 4 \pmod 8$.
This shows that the greatest power of $2$ that divides $19^{88}$ is $2 \times 2 \times 2 \times 4 = 2^5$.
Now for $q$, notice that:
$$19^{k}+1 \equiv 1^k+1\equiv 2 \pmod 3$$
the factor of $3^q$ comes from the factor $19^{11}-1$. Now we use the factorization:
$$19^{11}-1 = (19-1)(19^{10}+19^9+19^8+\dots +19 + 1)$$
The latter factor $\equiv 11 \pmod 3$. Hence the greatest power of $3$ that divides $19^{11}-1$ is $9$ (from $19-1=18$).
Hence the largest factor of $19^{88}-1$ of the form $2^p3^q$ is precisely $2^53^2$.
