let $\{p_1,p_2,p_2,\cdots ,p_r\}$ be the set of $r$($\ge2$) pair wise distinct prime numbers
i.e.., $(i\ne j \implies p_i \ne p_j)$ for all $1\le i,j\le r$
${Statement}$ :
For any such $\{p_1,p_2,p_2,\cdots ,p_r\}$ there exists a unique set of integers $\{l,l_1,l_2,l_3,\cdots, l_r \}$, satisfying
$\dfrac{1}{p_1p_2p_3\cdots p_r}+\dfrac{l_1}{p_1}+\dfrac{l_2}{p_2}+\dfrac{l_3}{p_3}+ \cdots + \dfrac{l_r}{p_r} = l$
where $1 \le l \le r-1$ and for all $1\le i \le r$ , $1 \le l_i \le p_i-1$
This follows from the theorem that an integer not divisible by a prime has a modular multiplicative inverse.
Write $P$ for the product of all primes in your set. Multiplying your equation by $P$ gives $lP=1+\sum_{i=1}^r l_iP/p_i$. Considering this equation modulo each of the primes gives $r$ congruences, $l_i\frac{P}{p_i}\equiv -1 \bmod p_i$. Each of these can be solved uniquely for an integer $1\leq l_i < p_i$ because $\frac{P}{p_i}$ is coprime to $p_i$, hence has an inverse $\bmod p_i$. The individual terms $l_i\frac{P}{p_i}$ are all positive, and strictly less than $P$, so the sum of $r$ of them (plus 1) is positive and strictly less than $rP$.
Hint $\ $ It follows exactly as in the Heaviside cover-up method for partial fractions. Given
$$\dfrac{1}{p_1p_2p_3\cdots p_r}+\dfrac{l_1}{p_1}+\dfrac{l_2}{p_2}+\dfrac{l_3}{p_3}+ \cdots + \dfrac{l_r}{p_r}\, =\, l$$
Multiplying the above by $\,p_r\,$ and evaluating mod $\,p_r\,$ yields
$$\dfrac{1}{p_1p_2p_3\cdots p_{r-1}} + l_r\, \equiv\, 0\pmod {p_r}$$
which determines $\,l_r\,$ uniquely, mod $\,p_r.\,$ (since the other $\,p$'s are coprime to $\,p_r\,$ they are invertible mod $\,p_r).\,$ Exactly the same works for the other $\,l_i,\, p_i.$ Note that one only needs that the $\,p_i\,$ are pairwise coprime (vs. distinct primes).
$r$= number of prime numbers in the set.
– hanugm Aug 11 '14 at 13:08