Problem 1050

Answer. No, it will not stop.

Solution. The condition can be rephrased as follows: if the number on the screen is of the form $100 A + B$, where $0 \leqslant B < 100$, then it is replaced by $2 A + 8 B$. Since $B < 100$, it follows that $8 B < 800$, so all numbers starting from 800 will necessarily decrease. A natural question is - can anything prevent the number from decreasing further? Or can we determine to what number, less than 800, the supercomputer will "count down"? First, from the above expression, it is clear that the difference between the number and its image is always divisible by 7. Thus, the remainder upon division by 7 is preserved. Since $1001 = 7 \cdot 11 \cdot 13$ is divisible by 7, so is $111111 = 111 \cdot 1001$, and therefore, the number consisting of $900 = 6 \cdot 150$ ones. This means that all resulting numbers will also be divisible by 7 - and by 14, because $2 A + 8 B$ is always even.

However, this is not enough. Does the divisibility by any other prime number $p$ remain unchanged (or change in a predictable way)?

If $100 A + B$ is divisible by $p$, then so is $8(100 A + B) = 800 A + 8 B$ (and if it is not divisible, the remainder is multiplied by 8). The difference between this number and $2 A + 8 B$ is $(800 - 2) A = 798 A$, so if $p$ is a divisor of 798, the remainder upon division by $p$ will be multiplied by 8. $798 = 7 \cdot 114 = 7 \cdot 2 \cdot 3 \cdot 19$; this gives us two new prime numbers, 3 and 19.

Let's see what remainder the initial number gives when divided by 3 and 19. It is divisible by 3 because 900 is divisible by 3. Finally, $10^{18} - 1$ is divisible by 19, which follows from Fermat's Little Theorem, but can also be verified by "long division". Therefore, the number consisting of $900 = 18 \cdot 50$ ones is divisible by 19. Hence, all iterations, starting from the second, will be divisible by 798. This means that the final (non-zero!) number will never be less than 100.

Comment. If we had not been "lucky" with the choice of the initial number - we could have "tracked" the changes in the remainders upon division by 3, 7, and 19 (since they are multiplied by 8 at each step), and determine whether there are any resulting numbers less than 100.