Question

${}^{\prime }{p}^{\prime }$ is prime ${}^{\prime }{n}^{\prime }$ is a positive integer and $n+p=2000.$

L.C.M. of $n$ and $p$ is $21879.$ Then find ${}^{\prime }{p}^{\prime }?$

Answer 1

Here, $p$ is prime number and $n$ is a positive integer.

$n+p=2000$ ----- ( 1 )

$np=21879$

$n=\frac{21879}{p}$ ---- ( 2 )

Substituting ( 2 ) in ( 1 ) we get,

$\Rightarrow$ $\frac{21879}{p}+p=2000$

$\Rightarrow$ $21879+p^2=2000p$

$\Rightarrow$ $p^2-2000p+21879=0$

$\Rightarrow$ $p^2-1989p-11p+21879=0$

$\Rightarrow$ $p(p-1989)-11(p-1989)=0$

$\Rightarrow$ $(p-1989)(p-11)=0$

$\Rightarrow$ $p=1989$ or $p=11$

It is given that $p$ is prime, so $p\ne 1989$

$\therefore$ $p=11$