For each positive integer k, define ak as the number obtained from adding k zeroes and a 1 to the right of 2024, all written in base 10. Determine wether there's a k such that ak has at least 20242024 distinct prime divisors.
Problem
Source: PErA 2024/6
Tags: number theory
04.03.2024 17:12
The answer is yes. I am gonna assume that ak=202400...00⏟k1=2024⋅10k+1+1. We use an induction on N that there exists k for which ak has at least N distinct prime divisors. The base case is trivial. For the induction step assume that we found some k such that ak has at least N distinct prime divisors and we will prove that there exists k′ for which ak′ has at least N+1 prime divisors. Let the primes p1<p2<⋯<pN be the divisors of ak=2024⋅10k+1+1. Define a sequence bm=2024⋅10mφ(2024⋅10k+1+1)+k+1 for m=1,2,3,⋯. Clearly, there are only finitely many primes dividing at least one of the terms of the sequence (bm)∞m≥1, therefore, using Kobayashi's theorem, there are infinitely many primes dividing at least one of the terms of the sequence cm=bm+1, for m=1,2,3,⋯. Choose a prime q>pN such that q∣ct for some t∈N. Now, notice that c_{t} = 2024 \cdot (10^{t})^{\varphi(2024 \cdot 10^{k+1} + 1)} \cdot 10^{k+1} + 1 \equiv 2024 \cdot 10^{k+1} + 1 \equiv 0 \pmod{2024 \cdot 10^{k+1} + 1} using that \gcd(10^t, 2024 \cdot 10^{k+1} + 1) = 1. What it means is that c_{t} is divisible by p_{1}, p_{2}, \cdots, p_{N} and also q, therefore if we choose k' = t \varphi({2024\cdot 10^{k+1} + 1}) + k, a_{k'} will have at least N+1 distinct prime divisors and we are done. \ \square