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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.

Jordan Glen
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ColdStorm
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    If $z=2$ and $z=3$ don't work, why doesn't option $(D)$? – J126 Apr 03 '15 at 11:49
  • with $z=2$ it' s impossible because if you solve the equation $$594=22(10+x)$$ you will find $x=17$ but $x$ must has one digit – dario Apr 03 '15 at 12:02

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We immediately get $zzv = 594-y74=?20$ from the addition part, so $z=2$ (and $v=0$ and $y=3$ and $wx=17$)

Hagen von Eitzen
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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.

SWilliams
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  • As you observed, $22 \cdot 17 = 374 \neq 594$. There is no valid solution. – N. F. Taussig Apr 03 '15 at 19:27
  • Adding 220 to your answer, which is a step in the problem, gets you to 594. Or did I miss something? – SWilliams Apr 03 '15 at 19:48
  • The factors are $10z + z$ and $10w + x$. If $z = 2$, $w = 1$, and $x = 7$, then $(10z + z)(10w + x) = 22 \cdot 17 = 374$. However, the final product is supposed to be $594$. – N. F. Taussig Apr 03 '15 at 19:57
  • There is an addition operation after the multiplication. – SWilliams Apr 03 '15 at 20:05
  • That is because the row $y74$ (meaning $100z + 70 + 4$) represents the product of $x$ and the number $10z + z$, while the row $zzv$ (meaning $100z + 10z + v$) represents the product of $10w$ and $10z + z$. To find the product $594 = (10w + x)(10z + z)$, we must add $x(10z + z)$ and $10w(10z + z)$. – N. F. Taussig Apr 03 '15 at 20:09
  • Hah! I get it now... just wasn't looking at it properly. Thanks! – SWilliams Apr 03 '15 at 20:12