New Year's Countdown Day 30: Goodbye 2017

Let $f(k) = \sqrt[30]{20x + k}.$ The equation in the problem can be rewritten as $f(f(f(17))) = 17.$ However, $f$ is a monotonically increasing function, so the only way that $f(f(f(k)))$ can equal $k$ is if $f(k) = k.$ Thus, the problem reduces to $f(17) = 17,$ and further manipulation gives us

$\begin{aligned} f(17) &= 17 \\ \sqrt[30]{20x + 17} &= 17 \\ 20x + 17 &= 17^{30} \\ x &= \dfrac{17^{30} - 17}{20}. \end{aligned}$

Therefore, the value of $N$ is $\dfrac{17^{30} - 17}{20},$ and $a + b + c = 17 + 30 + 20 = \boxed{67}.$

We can solve this by exploiting the structure of the equation. Suppose that we had an $x_0$ such that $20x_0 + 17 = 17^{30}$ . Then the innermost radical would evaluate to $17$ and again we would have $20x_0 + 17$ so then the second radical also reduces to $17$ and then for the last radical we also end up with $20x_0 + 17$ so we found a solution. And we may solve for this $x_0$ , getting $x = \frac{17^{30} - 17}{20}$ . To show this is the only solution first notice that for $[0, \infty]$ the function is increasing so no other positive solutions exist. Then, for the function to be defined for negative $x$ , it must be for really small $|x|$ , smaller than $\frac{17}{20}$ so a trivial size check for the function shows it won't reach 17.

The whole left hand side expression equals to 17, so, Substitute 17 in left hand side with the expression i.e., $\sqrt[30]{20x+\sqrt[30]{20x+\sqrt[30]{20x+\sqrt[30]{20x+\sqrt[30]{20x+\sqrt[30]{20x+17}}}}}} =17$

We can repeat the above logic to get an infinite expansion like this: $\sqrt[30]{20x+\sqrt[30]{20x+ \sqrt[30]{20x+ ..... \infty}}} = 17$ This implies: $\sqrt[30]{20x+17} = 17$ So, $x = \frac{17^{30} - 17}{20}$

There are no other real solutions to the equation, so,

$N=x=\frac{17^{30} - 17}{20}$ i.e., $a = 17, b = 30, c = 20$ , where $a$ and $c$ are coprimes.

So, $\boxed{a+b+c=67}$

Simply use the first and second binomial coefficients to manipulate the sum of roots.