Which of the following numbers is divisible by $3^{10} + 2$?

We need to check divisibility by $3^{10}+2$. Let’s denote

$N = 3^{10}+2$.

Check option 1: $3^{20}+4$

Write $3^{20}+4 = (3^{10})^2+4$.

Since $N=3^{10}+2$, substitute $3^{10} = N-2$.

So $3^{20}+4 = (N-2)^2+4 = N^2-4N+8$.

Clearly divisible by $N$? No, because remainder $= -4N+8$, not multiple of $N$.

So not divisible.

---

Check option 2: $3^{30}+2$.

$3^{30}+2 = (3^{10})^3+2 = (N-2)^3+2$.

Expand: $(N^3 - 6N^2 + 12N - 8)+2 = N^3 - 6N^2 + 12N - 6$.

This is divisible by $N$? Yes, because $N^3$, $-6N^2$, $12N$ are multiples of $N$, remainder $-6$.

So not divisible.

---

Check option 3: $3^{20}+8$.

$3^{20}+8 = (3^{10})^2+8 = (N-2)^2+8 = N^2-4N+12$.

Clearly divisible? Terms: $N^2-4N$ divisible, remainder $12$.

So not divisible.

---

Check option 4: $3^{30}+8$.

$3^{30}+8 = (3^{10})^3+8 = (N-2)^3+8 = (N^3-6N^2+12N-8)+8 = N^3-6N^2+12N$.

This equals $N(N^2-6N+12)$, which is divisible by $N$.

So divisible.

---

Answer: $3^{30}+8$