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Q. 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 $P_1, P_2,\ldots P_{2020}$ as $2020$ distinct primes, we can say there exists $n$, such that ${ n=0 (\mod P_1^{1398}), n=-1 (\mod P_2^{1398}),\ldots, n=-2019 (\mod P_{2020}^{1398}) }$ 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.)

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    $\begingroup$ For some basic information about writing mathematics at this site see, e.g., here, here, here and here. $\endgroup$ Commented May 12 at 17:51
  • $\begingroup$ Please add context per site rules. This reads like a contest question. Please say which contest this came from, or if it is not from a contest, where this question came from. Also, mathjax is your friend. $\endgroup$ Commented May 12 at 17:56
  • $\begingroup$ Hi. I don't know where the question is from. A friend of mine had this as homework a few months ago, and it wasn't solved in class, so I asked it here. $\endgroup$ Commented May 12 at 18:02
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    $\begingroup$ Your idea almost works. The only problem is that you might have not only $n\equiv 0\bmod P_1^{1398}$, but $n\equiv 0$ mod some higher power of $P_1$, and similarly for the other constraints. Can you modify your approach to compensate for that? $\endgroup$ Commented May 12 at 18:34
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    $\begingroup$ As the question states, the number of divisors of each number is a multiple of 1399, not necessarily 1399. $\endgroup$ Commented May 12 at 18:41

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As noted in the comments, the problem is that for every constraint you may have that your number is divisible by a higher power of each $P_i$; a way to fix this is the following. Consider the system of equations

\begin{cases}x\equiv -1+P_1^{1398}&\mod P_1^{1399}\\ x\equiv -2+P_2^{1398}&\mod P_2^{1399}\\&\vdots\\x\equiv -2020+P_{2020}^{1398}&\mod P_{2020}^{1399}\end{cases}

which by the Chinese remainder theorem has an integer solution $x$. Now for each $i\in[2020]$ we have that $x+i$ is divisible by $P_i^{1398}$ but not by $P_i^{1399}$, and so, the number of divisors of each $x+i$ is divisible by $1398+1=1399$ and trivially the $x+i$ form $2020$ consecutive numbers

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