Summing Prime Factors

As the number is symmetric so we can easily break it up into prime factors

Firstly, $99899 = 9\times 10^4 + 9\times 10^3 + 8\times 10^2 + 9\times 10^1 +9\times 10^0$

Now let $10 = x$

So, $99899 = 9x^4 + 9x^3 + 8x^2 + 9x + 9$

Dividing by $x^2$

$\rightarrow x^2\left(9x^2 + 9x + 8 + \dfrac9x + \dfrac{9}{x^2} \right)$

$\rightarrow x^2 \left(9\left(x^2 + \dfrac{1}{x^2} \right) + 9\left( x+\dfrac1x \right) + 8 \right)$

Let $\left(x+\dfrac1x \right) = t$ So, $\left(x^2 + \dfrac{1}{x^2} \right) = t^2 -2$

Substituting values:

$x^2 \left( 9t^2 + 9t -10 \right)$

Factorizing,

$x^2(3t-2)(3t+5)$

$\rightarrow x^2 \left(3\left(x+\dfrac1x \right)-2 \right) \left(3\left(x+\dfrac1x \right)+5 \right)$

$\rightarrow 100\times (30 + 0.3 -2 )(30 + 0.3 +5)$

$\Rightarrow 283 \times 353$

So, their sum is $283 + 353 = \boxed{636}$

${ 318 }^{ 2 }-{ 35 }^{ 2 }=99899\\ \Rightarrow \quad 283\cdot 353=99899\\ 283+353=636$