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Is there any way to get something out of: $50x + 5y - 18z = 0$ ?

I found this trying to solve this:

\begin{equation} xyz+xyz+xyz=zzz \end{equation}

in which xyz are not multiplying themselves. They're just variable of a number; for example, let's say that x=2;y=1;z=8, then xyz=218. As you can see in the example before, x is in the hundreds place, y in the tens and z in the ones.

Then:

\begin{equation} xyz + xyz + xyz = zzz \end{equation} \begin{equation} 3 (xyz) = zzz \end{equation} \begin{equation} 3 [(x*100)+(y*10)+(z*1)] = (z*111) \end{equation} \begin{equation} 50x + 5y - 18z = 0 \end{equation}

From here I couldn't get nothing more than separating x,y or z; nothing more than variables.

Is there any way to extract values from it? I was thinking about solving it by linear algebra, but it's only 1 equation (and 3 variables).


The answer is:

x=1; y=8; z=5


Thanks advance,
Luiz.

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    are $$x,y,z$$ numbers with $$0\le x,y,z\le 9$$? – 2017-02-19
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    sorry, I forgot to mention that. But as Rohan said above: 0 < x ≤ 9 and 0 ≤ y, z ≤ 9. – 2017-02-19
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    i have found $$x=1,y=8,z=5$$ solving this Diophantine equation – 2017-02-19
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    Thank you!! I wish I could vote your comment up, but I don't have enough reputation to do it. – 2017-02-19
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    i will waiting for the next question from you! – 2017-02-19

2 Answers 2

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Well, we need that $0 as they are all the digits in a three-digit number. Thus, we can rearrange our equation as: $$z = \frac {50x+5y}{18} = \frac {5 (10x+y)}{18}$$

It is obvious that the numerator's last digit is always $5$ or $0$, the former of which will not make the numerator to be divisible by $18$. So, the last digit of the numerator is $0$, and the first number satisfying this condition divisible by $18$ is $90$. Thus, $z=5$. Now, it will bring easy to see what values can be taken by $x $ and $y $.

Hope it helps.

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    I was just writing it down. Thank you for the answer!! – 2017-02-19
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    @Luiz You are welcome. (+1) to you! – 2017-02-19
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This question is a logic problem, rather than a straight algebra problem.

You have the equation $xyz+xyz+xyz=zzz$, where $0

Combining the left side gives $3(xyz)=zzz$. Expanding as you did gives $300x+30y+3z=100z+10z+z$.

Notice the last "digit" of each side. We need $3z=z$ thinking in terms of last digit. The only digits for which this is true are $z=0,5$.

If $z=0$, then you will find all digits must be 0. No good!

If $z=5$, then we have $3(xy5)=555$. Dividing by 3 gives $xy5=185$, which gives the solution.

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    Thank you!! Your answer answer helped as much as Rohan's, but as he answered before, I accepted his. I wish I could at least vote your answer up, but I don't have enough reputation. – 2017-02-19