Prove that there are 2020 consecutive numbers such that the number of divisors of each number is a multiple of 1399? I thought maybe choosing P1, P2,...P2020 as 2020 distinct primes, we can say there exists n, such that { n=0 (mod P1^1398), n=-1 (mod P2^1398) ... n=-2019 (mod P2020^1398) } with the Chinese remainder theorem, but I don't think this is right and I don't know the reason. If this statement is true, does it answer the question, and if not, how should I proceed? (Also, 1399 and 2020 aren't special numbers, so I suppose 1399 being a prime is an accident, and I shouldn't focus on it. But I don't know how I would think about the problem if the numbers differed and the number chosen instead of 1399 was composite.)