What would be the value of z in this question?
If $z=2,$ the relation becomes $22\cdot wx = 594,$ which gives $wx=27.$ Partial product of $22\cdot 27$ is $154 + 440.$ It's incongruous with the partial product given in the question.
$z=3$ is probably incorrect. So my question is: It $z=2$ a correct option? This is the image from an official answer key of some government exam uploaded on their website.
On
Let $v, w, x, y$, and $z$ be positive integers in $(0,...,9)$.
First, we have that $yz4 + zzv = 594$ which implies that $v=0$, $z=2$ and $y=3$ and that $yz4$ must be $374$. Next, since $z=2$, we have $22\times wx = 374$ which implies $wx=17$ which in turn implies $w=1$ and $x=7$.
So letting $v=0, x=7, w=1,$ and $z=2$ gives $22 \times 17 = 374$, Since $zzv$ must be $220$, then $374 + 220=594$.
So it looks like $z=2$ is the correct answer.
We immediately get $zzv = 594-y74=?20$ from the addition part, so $z=2$ (and $v=0$ and $y=3$ and $wx=17$)