$p$ is prime and $n$ is a positive integer and $n + p = 2000$. LCM of $n$ and $p$ is $21879$. Then find $n$.

Hint: If $n$ is a multiple of $p$, then the LCM of $n$ and $p$ will be $n$. When $n$ and $p$ are relative prime numbers, then the LCM of $n$ and \[p\] will be $np$.
Later find the value of p in terms of n and substitute in the equation given in the question. Solving the equation we can find the value of n.

Complete step-by-step answer:
We know that if $n$ is a multiple of $p$, then the LCM of $n$and $p$will be $n$.
Thus, $n = 21879$
This is not possible because $n + p = 2000$
Thus, $n$ and $p$ are relative prime numbers.
So, LCM of $n$ and \[p\] will be $np$.
Thus, $np = 21879$
$ \Rightarrow p = \dfrac{{21879}}{n}$..........….. (1)
We have, $n + p = 2000$..........….. (2)
$ \Rightarrow n + \dfrac{{21879}}{n} = 2000$ [Using(1)]
$ \Rightarrow \dfrac{{{n^2} + 21879}}{n} = 2000$
$ \Rightarrow {n^2} + 21879 = 2000n$
$ \Rightarrow {n^2} - 2000n + 21879 = 0$..........….. (3)
Now use factoring method to solve the above quadratic equation,
$ \Rightarrow {n^2} - \left( {11 + 1989} \right)n + 21879 = 0$
$ \Rightarrow {n^2} - 11n - 1989n + 21879 = 0$
$ \Rightarrow n\left( {n - 11} \right) - 1989\left( {n - 11} \right) = 0$
$ \Rightarrow \left( {n - 11} \right)\left( {n - 1989} \right) = 0$
$ \Rightarrow n = 11$ or $n = 1989$
Thus, \[p = 2000 - n\] [Using(2)]
$p = 2000 - 11$ or $p = 2000 - 1989$
$p = 1989$ or $p = 11$
Since, $p$ is prime, so $p$ cannot be $1989$.

Thus, $n = 1989$ and $p = 11$.

Note: The quadratic equation (3) can also be solved by the quadratic formula which is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Compare the quadratic equation (3), i.e., ${n^2} - 2000n + 21879 = 0$ with the standard quadratic equation $a{x^2} + bx + c = 0$, we get-
$a = 1,b = - 2000,c = 21879$ and $x = n$
Now, $n = \dfrac{{ - \left( { - 2000} \right) \pm \sqrt {{{\left( { - 2000} \right)}^2} - 4 \times 1 \times 21879} }}{{2 \times 1}}$
$ \Rightarrow n = \dfrac{{2000 \pm \sqrt {4000000 - 87516} }}{2}$
$ \Rightarrow n = \dfrac{{2000 \pm \sqrt {3912484} }}{2}$
$ \Rightarrow n = \dfrac{{2000 \pm 1978}}{2}$
$ \Rightarrow n = \dfrac{{2000 + 1978}}{2}$ or $n = \dfrac{{2000 - 1978}}{2}$
$ \Rightarrow n = \dfrac{{3978}}{2}$ or $n = \dfrac{{22}}{2}$
$ \Rightarrow n = 1989$ or $n = 11$