Question

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

Answer 1

If $n$ is a multiple of $p$, then L.C.M of $n$ and $p$ will be $n$.
Thus, $n=21879$
This is not possible as $n+p=2000$
Thus, $n$ and $p$ are relatively prime numbers.
So, L.C.M of $n$ and $p$ will be $np$
Thus, $np=21879$ ----------(i)
Now, $n+p=2000$
$=>n+\frac{21879}{n}=2000$ -----(using (i))
$=>\frac{n^2+21879}{n}=2000$
$=>n^2+21879-2000n=0$
$=>n^2-2000n+21879=0$
$=>n^2-11n-1989n+21879=0$
$=>n(n-11)-1989(n-11)=0$
$=>(n-11)(n-1989)=0$
Thus, $n=11$ or $n=1989$
Thus, $p=2000-11$ or $p=2000-1989$
$p=1989$ or $p=11$
$p$ is prime, so $p$ cannot be $1989$
Thus, $n=1989$ and $p=11$