Prove by mathematical induction that for all positive integers, n, 10ⁿ + 3 x 4^(n+1) + 5 is divisible by 3.

Let $P(n)$ be the statement $10^n + 3 \times 4^{n+1} + 5$ is divisible by 3.
For $n=1$, we evaluate the expression:

$$10^1 + 3 \times 4^{1+1} + 5$$ $$= 10 + 3 \times 4^2 + 5$$ $$= 10 + 3 \times 16 + 5$$ $$= 10 + 48 + 5$$ $$= 63$$

Since $63 = 3 \times 21$, 63 is divisible by 3.
Thus, $P(1)$ is true.

Assume that $P(k)$ is true for some positive integer $k$.
This means $10^k + 3 \times 4^{k+1} + 5$ is divisible by 3.
We can write this as:

$$10^k + 3 \times 4^{k+1} + 5 = 3m$$

for some integer $m$.

We need to show that $P(k+1)$ is true, i.e., $10^{k+1} + 3 \times 4^{(k+1)+1} + 5$ is divisible by 3.
Consider the expression for $P(k+1)$:

$$10^{k+1} + 3 \times 4^{k+2} + 5$$

We can rewrite $10^{k+1}$ as $10 \times 10^k$ and $4^{k+2}$ as $4 \times 4^{k+1}$:

$$10 \times 10^k + 3 \times 4 \times 4^{k+1} + 5$$ $$= 10 \times 10^k + 12 \times 4^{k+1} + 5$$

From the inductive hypothesis, we know $10^k = 3m - 3 \times 4^{k+1} - 5$. Substitute this into the expression:

$$= 10(3m - 3 \times 4^{k+1} - 5) + 12 \times 4^{k+1} + 5$$ $$= 30m - 30 \times 4^{k+1} - 50 + 12 \times 4^{k+1} + 5$$

Combine like terms:

$$= 30m - (30 - 12) \times 4^{k+1} - (50 - 5)$$ $$= 30m - 18 \times 4^{k+1} - 45$$

Factor out 3 from each term:

$$= 3(10m - 6 \times 4^{k+1} - 15)$$

Since $m$ is an integer and $k$ is a positive integer, the expression $(10m - 6 \times 4^{k+1} - 15)$ is an integer.
Therefore, $10^{k+1} + 3 \times 4^{k+2} + 5$ is a multiple of 3, which means it is divisible by 3.
Thus, $P(k+1)$ is true.

By the principle of mathematical induction, the statement $10^n + 3 \times 4^{n+1} + 5$ is divisible by 3 for all positive integers $n$.