i need help

1.

Obviously, $n\geq 7$.

$\quad 235236_n\\ =2n^5 + 3n^4 + 5n^3 + 2n^2 + 3n + 6$ 

This holds true for any $n\geq 7$.

Then, clearly, $2n^5+3n^4+5n^3+2n^2+3n\equiv 1\pmod7$.

Define $P(n) = 2n^5+3n^4+5n^3+2n^2+3n-1$.

Consider all possible values of P(n) mod 7 when $0 \leq n \leq 6, n\in \mathbb Z$.

$P(0) \equiv 6\pmod 7\\ P(1) \equiv 0 \pmod 7\\ P(2) \equiv 4 \pmod 7\\ P(3) \equiv 1 \pmod 7\\ P(4) \equiv 1 \pmod 7\\ P(5) \equiv 1\pmod 7\\ P(6) \equiv 1 \pmod 7\\ $

Generally, $P(7k + 1) \equiv 0 \pmod 7 \quad\forall \;k\in \mathbb Z^+$.

All possible values of n are: 8,15,22,29,36,43,50,57,64,71,78,85,92,99.

So there are 14 possible values of n such that $235236_n$ is divisible by 7.

2.

$AA_5 = 5A + A = 6A \quad \forall \; A \in [0,4] \land A\in \mathbb Z\\ BB_7 = 7B + B = 8B \quad \forall \; B \in [0,6] \land B\in \mathbb Z$

From this, we know that the required base-10 integer must be divisible by both 6 and 8.

Least common multiple of 6 and 8 = 24.

So the answer is 24.

Check: $24_{10} = 44_5 = 33_7$