$$25^{190} - 19^{190} - 8^{190} + 2^{190}$$ is divisible by

We need to determine whether $$25^{190} - 19^{190} - 8^{190} + 2^{190}$$ is divisible by 14, 34, both, or neither.

Note that $$14 = 2 \times 7$$ and $$34 = 2 \times 17$$. So we need to check divisibility by 2, 7, and 17.

To begin,

$$25^{190}$$ is odd, $$19^{190}$$ is odd, $$8^{190}$$ is even, $$2^{190}$$ is even.

So $$25^{190} - 19^{190} - 8^{190} + 2^{190} = (\text{odd} - \text{odd}) - (\text{even} - \text{even}) = \text{even} - \text{even} = \text{even}$$.

The expression is divisible by 2.

Next,

Regroup: $$(25^{190} - 8^{190}) - (19^{190} - 2^{190})$$.

Since $$25 - 8 = 17$$, by the algebraic identity $$a^n - b^n$$ is divisible by $$(a - b)$$, we get $$17 \mid (25^{190} - 8^{190})$$.

Since $$19 - 2 = 17$$, similarly $$17 \mid (19^{190} - 2^{190})$$.

Therefore $$17 \mid \left[(25^{190} - 8^{190}) - (19^{190} - 2^{190})\right]$$.

From this,

Since the expression is divisible by both 2 and 17, and $$\gcd(2, 17) = 1$$, the expression is divisible by $$2 \times 17 = 34$$.

Continuing,

We check modulo 7:

$$25 \equiv 4 \pmod{7}$$, $$19 \equiv 5 \pmod{7}$$, $$8 \equiv 1 \pmod{7}$$, $$2 \equiv 2 \pmod{7}$$.

By Fermat's little theorem, $$a^6 \equiv 1 \pmod{7}$$ for $$\gcd(a, 7) = 1$$.

Since $$190 = 6 \times 31 + 4$$, we have $$a^{190} \equiv a^4 \pmod{7}$$.

$$4^4 = 256 \equiv 4 \pmod{7}$$

$$5^4 = 625 \equiv 2 \pmod{7}$$

$$1^4 = 1 \pmod{7}$$

$$2^4 = 16 \equiv 2 \pmod{7}$$

So the expression $$\equiv 4 - 2 - 1 + 2 = 3 \pmod{7}$$.

Since $$3 \not\equiv 0 \pmod{7}$$, the expression is not divisible by 7, and hence not divisible by 14.

Conclusion: The expression is divisible by 34 but not by 14.

The correct answer is Option C.