New Year's Countdown Day 2: Two to 2x

Let $N = \overline{a_n a_{n - 1} a_{n - 2} \dots a_2 a_1},$ where all the $a_i$ s are digits such that $a_n, a_2 \neq 0$ (to avoid leading zeroes). We want to find the smallest value of $N$ such that

$2 \overline{a_n a_{n - 1} a_{n - 2} \dots a_2 a_1} = \overline{a_2 a_1 a_n \dots a_4 a_3}.$

Let $A = \overline{a_n a_{n - 1} \dots a_4 a_3}$ and $B = \overline{a_2 a_1}.$ The condition rewrites and simplifies into

$\begin{aligned} 2(100A + B) &= 10^{n - 2}B + A \\ 199A &= (10^{n - 2} - 2)B. \end{aligned}$

The right hand side of the equation must be divisible by 199. Since 199 is prime, we must have either $199|10^{n - 2} - 2$ or $199|B.$ As $B$ has two digits, it cannot be divisible by a three-digit number, so it must be true that $199|10^{n - 2} - 2.$ Writing this in modulo notation yields

$\begin{aligned} 10^{n - 2} - 2 &\equiv 0 \! \! \! \! \pmod{199} \\ 10^{n - 2} &\equiv 2 \! \! \! \! \pmod{199} \\ 100(10^{n - 2}) &\equiv 100 \cdot 2 \! \! \! \! \pmod{199} \\ 10^n &= 1 \! \! \! \! \pmod{199}. \end{aligned}$

Since we wish to minimize $N,$ we want the smallest value of $n$ for which this is true. But that is precisely equal to $\text{ord}_{199}(10),$ which we can calculate to be $n = 99.$ Thus, $N$ has 99 digits, while $A$ has 97 digits.

Plugging this value of $n$ back into our original condition yields

$\begin{aligned} 199A &= (10^{97} - 2)B \\ A &= \dfrac{10^{97} - 2}{199}B. \end{aligned}$

The number $\dfrac{10^{97} - 2}{199}$ has $\left \lfloor \log_{10} \left ( \dfrac{10^{97} - 2}{199} \right) + 1\right \rfloor = 95$ digits, so we seek the smallest value of $B$ that will make $A$ have 97 digits. By inspection, the minimum such value of $B$ is 20, giving us $A = \dfrac{20(10^{97} - 2)}{199}$ and $N = \dfrac{20(10^{97} - 2)}{199} \cdot 10^2 + 20 = \dfrac{20(10^{99} - 1)}{199}.$

To find the value of $k,$ we find the last three digits of the value of $N$ we just found. The last two digits are $B = 20.$ The hundreds place is the units digit of $A,$ which is 0 because of the factor of 20 in its value. Thus, the last three digits of $N$ are $k = 020 = 20.$

Our final answer is $nk = 99(20) = \boxed{1980}.$ The value of $N$ written out in full is

$N = 100,502,512,562,814,070,351,758,793,969,849, \\ \quad 246,231,155,778,894,472,361,809,045,226,130, \\ \quad 653,266,331,658,291,457,286,432,160,804,020.$