Question

Among three co-prime numbers, the product of the first two is $18304$ and the product of the last two is $72576$, find the sum of these numbers.

• A. $588$
• B. $764$
• C. $686$
• D. $838$
• E. $Noneofthese$

Answer 1

The correct option is D $838$

Let three co-prime numbers be $x,y$ and $z$

We have $xy=18304$ and $yz=72576$

$HCF$ of $18304$ and $72576$ is

$18304)72576(3-54912------------17664)18304(1-17664-------------640)17664(27-1280-----------4864-4480-----------384)640(1-384---------256$

$256)384(1-256------------128)256(2-256-----------x$

$\therefore HCF=128⟶$ second number $\left(y\right)$

$\therefore x=\frac{18304}{128}=143\to$ first number.

$z=\frac{72576}{128}=567\to$ third number.

$\therefore$ sum of three numbers $=x+y+z=128+143+567$

$=838$