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

Guest Aug 26, 2023

#1**0 **

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 \(\boxed{9}\).

SpectraSynth Aug 26, 2023