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# Positive integers

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If x, y, and z are positive integers such that 6xyz + 30xy + 18xz + 12yz + 10x + 17y + 5z =481, find x + y + z.

Aug 26, 2023

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Given the equation $$6xyz + 30xy + 18xz + 12yz + 10x + 17y + 5z = 481$$, we want to find positive integer solutions for $$x$$, $$y$$, and $$z$$ such that the equation holds.

Notice that all the terms on the left-hand side are divisible by $$5$$, so let's divide the entire equation by $$5$$:

$5(2xyz + 6xy + 3xz + 2yz) + 2x + 17y + z = 96.$

Now, we can see that the left-hand side is a sum of terms, each of which is divisible by $$5$$, except for $$2x$$, $$17y$$, and $$z$$.

For the equation to hold, the left-hand side must be a multiple of $$5$$, and that means $$2x + 17y + z$$ must be a multiple of $$5$$.

Looking at the possible values of $$2x + 17y + z$$, we notice that $$2x + 17y + z$$ ranges from $$2 + 17 + 1 = 20$$ (when $$x = 1$$, $$y = 1$$, $$z = 1$$) to $$2 \cdot 6 + 17 \cdot 2 + 6 = 56$$ (when $$x = 6$$, $$y = 2$$, $$z = 6$$).

Since $$20 \leq 2x + 17y + z \leq 56$$, and we want it to be a multiple of $$5$$, the only possibility is $$25$$ (which is the only multiple of $$5$$ in that range).

So, we must have $$2x + 17y + z = 25$$.

To find the positive integer solutions for $$x$$, $$y$$, and $$z$$ that satisfy this equation, we can try different values for $$x$$, $$y$$, and $$z$$ in a systematic way. Starting from $$x = 1$$, we can substitute values for $$y$$ and $$z$$ that satisfy the equation $$17y + z = 25 - 2x$$.

We find that $$x = 1$$, $$y = 1$$, and $$z = 7$$ satisfy the equation $$2x + 17y + z = 25$$.

Therefore, $$x + y + z = 1 + 1 + 7 = 9$$.

So, the value of $$x + y + z$$ is $$\boxed{9}$$.

Aug 26, 2023