There is some natural number $n>1$ on the board. Operation is adding to number on the board it maximal non-trivial divisor. Prove, that after some some operations we get number, that is divisible by $3^{2000}$ A. Golovanov
Problem
Source: Tuymaada 2015, Day 2, Problem 5, Senior League
Tags: number theory
Vietnamisalwaysinmyheart
10.07.2017 16:25
Here is my solution:
Denote $d(n)$ is the largest non-trivial divisor of $n$
Claim: after some operation, there must have a number which is divisible by $2$ on the board.
Proof: If $n$ is odd then $n+d(n)$ is even. $\blacksquare$
Claim, after some operation, there must have a number which is divisible by $3$ on the board.
Proof: Now, taking an even number: $2n$ onboard, the next number will be write is $3n$ $\blacksquare$
Now, taking the number which is divisible by $3$ onboard, we have two cases:
If it has form of: $3(2k+1)$, We will prove by induction that, if onboard, there is number: $3^n(2k+1)$ then there will be $3^{n+1}(2k+1)$. this is, of course, after 3 operations, we have: $3^n(2k+1)\rightarrow (3^n+3^{n-1})(2k+1\rightarrow (3^n+3^{n-1}+\frac{3^n+3^{n-1}}{2})(2k+1)=3^{n+1}(2k+1)$, which is done.
If it has form of: $3.2^s(2k+1)$ with positive integer $s$, we have theses operation:
$3.2^s(2k+1)\rightarrow 3(2^s+2^{s-1})(2k+1))=3^2.2^{s-1}(2k+1)\rightarrow...\rightarrow 3^{s+1}(2k+1) $, by above induction, we are done