Let $n>1$ be a positive integer and $d_1<d_2<\dots<d_m$ be its $m$ positive divisors, including $1$ and $n$. Lalo writes the following $2m$ numbers on a board: \[d_1,d_2\dots, d_m,d_1+d_2,d_2+d_3,\dots,d_{m-1}+d_m,N \]where $N$ is a positive integer. Afterwards, Lalo erases any number that is repeated (for example, if a number appears twice, he erases one of them). Finally, Lalo realizes that the numbers left on the board are exactly all the divisors of $N$. Find all possible values that $n$ can take.
Problem
Source: 2022 Mexican Mathematics Olympiad P5
Tags: number theory
Justpassingby
11.11.2022 09:12
I proposed this problem : ) There are some complicated ways to do this problem, and some very simple ways. I feel like its pretty successful in that very simple ideas lead to simple solutions, but they're hard to find, and its actually doable with more involved techniques if one has enough tenacity.
starchan
11.11.2022 11:40
The answer is all powers of $2$.
Let $p$ be the smallest prime dividing $n$. Note that then this means $n(p+1) \mid Np$ and $n \mid N$ leading to $n(p+1) \mid N$ from where it follows that $N = kn(p+1)$ for a positive integer $k$. Let $q$ be the smallest prime dividing $N$. The largest number written on the board other than $N$ must be $N/q$, so $N/q = n(1+1/p)$ which implies $q = kp$. Since both $p, q$ are primes we obtain $p = q$ and $k = 1$. Thus $N = n(p+1)$. It's clear that $p = 2$ now, since otherwise $N$ is even and $q = 2 \ne p$, a contradiction. At this point we are done. We claim that all the solutions are $(n, N) = (2^u, 3 \cdot 2^u)$ for positive integers $u$. Suppose that $p$ is the smallest odd prime dividing $n$. Let $i$ be such that $d_i = p$. Note that then $d_{i-1} = 2^v$ for some $v > 0$. Now we know that $\frac{n}{p} + \frac{n}{2^v} \mid 3n$ from where it follows that $p+2^v \mid 3$ which is impossible since the left hand side always exceeds $3$. Hence such a $p$ does not exist and we have $n = 2^u$. It is easy to see that all such $n$ work.
emi3.141592
28.10.2024 21:19
Error.........