We have distinct arithmetic progressions with common differences $d_1, d_2, \cdots,d_n$. If every natural number belongs to exactly one arithmetic progression, prove that $\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_n}=1$.
I saw two proofs of this but none of them explained it clearly. The first one was to choose a large number $k$ and deduced that the numbers not greater than $k$ appearing in $i$th sequence is $\frac{k}{d_i}+/-$ a constant. I didn't understand what they did here. Finally they added all such numbers and got $\frac{k}{d_1}+\frac{k}{d_2}+\cdots+\frac{k}{d_n}+$ a constant$=k$ and by choosing $k$ large enough they said $\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_n}=1$ which i didn't get how.
Next they used a density argument saying that the density of first A.P is $\frac{1}{d_1}$ and such which also i didn't understand. It would be very helpful if these solutions are explained in simple language.
You don't quote or link to the arguments you've seen, so I'll try to spell out what they likely meant.
Of the positive integers from $1$ to $k$ inclusive (hereafter "small" numbers), the number in the progression of common difference $d_i$ can be bounded as follows. Write $k=qd_i+r$ for integers $q,\,r$, specified uniquely by $0\le r<d_i$. If some integer from $1$ to $r$ inclusive is in the progression, $q+1$ small numbers are in the progression; if not, only $q$ are. (It can't be even fewer, since this would require the least small number in the progression to exceed $d+r$, even though we could subtract $d$ repeatedly to obtain a counterexample.) Since$$qd_i\le k<(q+1)d_i,$$$q>k/d_i-1$ and $q+1\le k/d_i+1$, so $k/d_i$ is within $1$ of the number of small numbers in the progression. Therefore, $|\sum_i k/d_i-k|\le n$, i.e. $|\sum_i 1/d_i-1|\le n/k$. This is true for all $k\in\Bbb N$, so $\sum_i 1/d_i=1$.