Question

If $x4y5z$ is exactly divisible by $9,$ then the least value of $(x+y+z)$ is

(a) $3$

(b) $6$

(c) $9$

(d) $0$

Answer 1

The correct option is (d): $0$

$\because$ $x4y5z$ is exactly divisible by $9$

$\therefore$ The sum of its digits must be divisible by $9$

$\therefore$ $x+4+y+5+z$ or $9+(x+y+z)$ must be divisible by $9$

Hence, Least value of $x+y+z=0$ as $9+0=9$ is divisible by $9$

Check:

When $x+y+z=0,$ means $x=0,$ $y=0,$ $z=0$ [We are taking only positive numbers]

When we put $x=0,y=0,$ and $z=0$ in $x4y5z,$ then

The resulting number will be $04050=4050$

And, $4050$ is divisible by $9$