Problem 1363

Since $19$ is a prime, the greatest proper divisor of $N = 19^{19}$ is $19^{18}$ and the first number obtained is $N_1 = 19^{19} - 19^{18} = 18 \cdot 19^{18}$. Note that if a number $m$ is even then the operation gives $m/2$. So the next number is $N_2 = N_1/2 = 9 \cdot 19^{18}$. The greatest proper divisor of $9 \cdot 19^{18}$ is $3 \cdot 19^{18}$, hence

$N_3 = 9 \cdot 19^{18} - 3 \cdot 19^{18} = 6 \cdot 19^{18}$.

Now, as it is an even number, $N_4 = 3 \cdot 19^{18}$. Its greatest proper divisor is $19^{18}$, then $N_5 = 3 \cdot 19^{18} - 19^{18} = 2 \cdot 19^{18}$ and $N_6 = 19^{18}$.

We see that $6$ operations transform $19^{19}$ to $19^{18}$ so it takes another $6$ to reach $19^{17}$, and so on. In order to end with $1 = 19^0$, the process takes up $6 \cdot 19 = 114$ operations.