The number of real solutions of $$x^7 + 5x^3 + 3x + 1 = 0$$ is equal to ______

We need to find the number of real solutions of $$x^7 + 5x^3 + 3x + 1 = 0$$. Let $$f(x) = x^7 + 5x^3 + 3x + 1$$, so $$f'(x) = 7x^6 + 15x^2 + 3$$. Since $$7x^6 \geq 0$$ and $$15x^2 \geq 0$$ for all real $$x$$ and $$3 > 0$$, it follows that $$f'(x) \geq 3 > 0$$ for all $$x \in \mathbb{R}$$, and hence $$f(x)$$ is strictly increasing on $$\mathbb{R}$$.

As $$x \to -\infty$$, $$f(x) \to -\infty$$ (dominated by $$x^7$$), and as $$x \to +\infty$$, $$f(x) \to +\infty$$. Since $$f$$ is continuous and strictly increasing, by the Intermediate Value Theorem it crosses the $$x$$-axis exactly once.

The equation $$x^7 + 5x^3 + 3x + 1 = 0$$ has exactly 1 real solution. Therefore, the correct answer is Option B: $$1$$.