Number theory/series olympiad question

Question

European Mathematical Cup 2024 (Ended yesterday and as per the rules discussion is now allowed)

Problem 1 Wiske wrote a 2024-digit positive integer on the blackboard. In each round of the game she erases the last digit of the number, let this be d, and writes down the sum of the remaining number and 2d in place of the old number. She repeats the same steps with the newly obtained number. After a certain number of rounds, Wiske found that the new number obtained was the same as the number in the last round and she stopped the game. What is the smallest possible 2024-digit integer that Wiske could have started with in this game?

My solution was the smallest multiple of $19$ greater than $10^{2023}$. Later at home I used a C program to compute the Modulo pattern and found that the answer is $100...004$.

The way I got my answer is by realizing that the only case where Old Number $=$ New Number $+\ 2d$ is when the Old Number is 19 as $19=1+2*9$. I then computed the numbers that result in 19 and found they are all multiples of 19, and then I tested a few of those and they were also all multiples of 19 and so I came to the half guessed conclusion that it must be a sequence of multiples of 19.

I tested this fact for some numbers with a C program and found it to be true, but $10^{2023}$ is just too large to check. And so I am wondering if my solution is even true, and if it is why is it true?

Answer 1

This is not by any means an elegant solution, but it's the first one I thought of and probably the one that requires the least thinking.

Let $n_0$ be the initial $2024$-digit number and $n_k$ be the number after the operation has been applied $k$ times. Some things to note:

$$ \begin{align*} d &= n_k \textrm{ mod } 10 \\ \textrm{Remaining Number After Erasure} &= \left\lfloor\frac{n_k}{10}\right\rfloor \\ a \textrm{ mod } b &= a-b\left\lfloor\frac{a}{b}\right\rfloor \end{align*} $$

With this in mind, we can say that: $$ \begin{align} n_{k+1} &= \textrm{Remaining Number After Erasure} + 2d \\ &= \left\lfloor\frac{n_k}{10}\right\rfloor + 2(n_k \textrm{ mod } 10) \\ &= \left\lfloor\frac{n_k}{10}\right\rfloor + 2n_k - 20\left\lfloor\frac{n_k} {10}\right\rfloor \\ &= 2n_k - 19\left\lfloor\frac{n_k}{10}\right\rfloor \end{align} $$ First, we want to show that $19$ is the only valid solution for $n_{k+1}=n_k$. Setting $n_{k+1}=n_k$, we have: $$ \begin{align} n_k - 19\left\lfloor\frac{n_k}{10}\right\rfloor &= 0 \\ \frac{n_k}{19} &= \left\lfloor\frac{n_k}{10}\right\rfloor \end{align} $$ For which it's pretty clear that $n_k=19$ is the only solution.

Now let's show that this operation will only produce a multiple of 19 if the input is also a multiple of $19$. If we instead set $n_{k+1}=19m$ for some positive integer $m$ we have: $$ \begin{align} 19\left(m+\left\lfloor\frac{n_k}{10}\right\rfloor\right) &= 2n_k \\ m+\left\lfloor\frac{n_k}{10}\right\rfloor &= \frac{2n_k}{19} \\ \end{align} $$ Where we see that $19 \mid 2n_k$ and so $19 \mid n_k$.

Now we have to show $n_{k+1} < n_k$ for all $19 \mid n_k$ and $n_k > 19$. This will imply that every multiple of $19$, when repeatedly plugged into this operation, will eventually reach $19$ itself. Set $n_k=19m$ where $m > 1$: \begin{align} n_{k+1} &= 38m - 19\left\lfloor\frac{19m}{10}\right\rfloor \\ &= 38m - 19\left\lfloor m+\frac{9m}{10}\right\rfloor \\ &= 19m - 19\left\lfloor\frac{9m}{10}\right\rfloor < n_k \end{align}

So you are correct in that $n_0$ must be the smallest multiple of $19$ greater than $10^{2023}$. For actually finding that value, we can use the following rule of modular arithmetic:

If $a \equiv b \pmod{\phi(m)}$ and $\gcd(n, m)=1$ then $n^a \equiv n^b \pmod{m}.$

($\phi(m)$ is Euler's Totient Function)

Since $\phi(19)=18$ and $2023 \equiv 7 \pmod{18}$, we have $10^{2023} \equiv 10^7 \equiv 15 \pmod{19}$. Finding $10^7 \textrm { mod } 19$ is a bit annoying, but very doable. And so, to get the smallest multiple of $19$, you just have to add $4$ to get $10^{2023} + 4$.

Answer 2

Let's denote by "$\operatorname{reduce}$" the function that maps a natural number $n$ to $n'+2d$, where $n'$ is the number $n$ with the last digit erased, and $d$ is the last digit of $n$.

Now, obviously we have $10n'+d \equiv n \pmod{19}$, and from this it follows by multiplying by $2$ that $n'+2d \equiv 2n \pmod{19}$. So we conclude that $$\operatorname{reduce}(n) \equiv 2n \pmod{19}.$$ In other words, the function $\operatorname{reduce}$ maps multiples of $19$ to multiples of $19$, and non-multiples of $19$ to non-multiples of $19$. I think this more or less shows what you want (you still have to show that every multiple of $19$ eventually gets you down to $19$ itself, but I'm assuming this will follow by taking some easy bounds).