An algebra problem by Kevin Li

We will first call n. n=4x=5y=9z To find the least possible value, we must find the least possible value of n, which is the least common multiple of 4, 5, and 9, which is 180. Therefore, we get 180=4x=5y=9z. Therefore, x=45, y=36, and z=20. 45+36+20=101 Therefore, the least possible value of x+y+z is 101.

The $LCD$ of $4,5,9$ is $180$ . Just divide $180$ by the the three numbers to get $x,y,z$ and add them.

$4x=5y=9z$

x, y, and z are positive integers.

$\implies x=45,y=36,z=20$

$\implies x+y+z=\boxed{101}$

If 4x=5y=9z we must find the smallest possible number divisible by 4,5&9

LCM(4,5,9)=180

180=4x=5y=9z. So (x,y,z)=(45,36,20)

x+y+z=101

To get the minimum values of x, y, and z, we need to find the LCM of 4, 5 and 9. Now LCM of 4, 5 and 9 is 180 Therefore x= 180÷4=45 y= 180÷5=36 And. z=180÷9=20 Therefore minimum sum of x, y and z is 101.