Inspired by " Nothing to be confused here " Part 2

We need to find $(f(23)-f(11)) \bmod 100 = f(23) \bmod 100 - f(11)\bmod 100$ . Let us consider $f(23)$ and $f(11)$ separately.

Since $\gcd(23, 100) = 1$ , we can use Euler's theorem and Carmichael's lambda function $\lambda (\cdot)$ . Note that $\lambda (100) = 20$ and $\lambda (20) = 4$ .

$\begin{aligned} f(23) & \equiv 23^{23^{23^{23^{23}} \bmod 4} \bmod 20} \text{ (mod 100)} \\ & \equiv 23^{23^{(24-1)^{23^{23}} \bmod 4} \bmod 20} \text{ (mod 100)} \\ & \equiv 23^{23^3 \bmod 20} \text{ (mod 100)} \\ & \equiv 23^{(20+3)^3 \bmod 20} \text{ (mod 100)} \\ & \equiv 23^7 \text{ (mod 100)} \\ & \equiv 23^2(20+3)^5 \text{ (mod 100)} \\ & \equiv 529(3^5) \text{ (mod 100)} \\ & \equiv 29(243) \text{ (mod 100)} \\ & \equiv 47 \text{ (mod 100)} \end{aligned}$

Let $n = 11^{11^{11^{11}}}$ . Then

$\begin{aligned} f(11) & \equiv 11^n \text{ (mod 100)} \\ & \equiv (10+1)^n \text{ (mod 100)} \\ & \equiv \left(10^n + 10^{n-1} + \cdots + \frac {100n(n-1)}2 + 10n + 1\right) \text{ (mod 100)} \\ & \equiv \left(0 + 0 + \cdots + 0 + 10\blue n + 1\right) \text{ (mod 100)} & \small \blue{\text{Note that }n \text{ ends with 1}} \\ & \equiv 10 + 1 \equiv 11 \text{ (mod 100)} \end{aligned}$

Therefore $f(23) - f(11) \equiv 47 - 11 \equiv 36 \text{ (mod 100)}$ $\implies \sqrt N = \sqrt{36} = \boxed 6$ .