Question

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

Answer 1

what a load of rubbish.
Clearly, if x=y=z=1, then
6xyz + 30xy + 18xz + 12yz + 10x + 17y + 5z = 6+30+18+12+10+17+5 = 98, not 481
I do not think there is any solution with positive values.

Answer 2

Factoring out x, y, and z from the given expression, we have $$6xyz + 30xy + 18xz + 12yz + 10x + 17y + 5z = (6yz+10)x +(30y+17)x + (18z+12)y + 5z = x(6yz+10) + y(30x+17) + z(18x+12) + 5z.$$Therefore, $x(6yz+10) + y(30x+17) + z(18x+12) + 5z = 481$. Since x, y, and z are positive integers, we need $6yz+10 = 30x+17 = 18xz+12 = 5z=1$. We see that $6yz+10 = 30x+17 = 5z=1$ leads to $(x,y,z) = (1,1,1)$. Therefore, $x+y+z = 1+1+1 = \boxed{3}$.

Answer 3

Apologies for the mistake in my previous response. You are correct, there doesn't seem to be any solution with positive values for x, y, and z.

Let's analyze the equation $$6xyz + 30xy + 18xz + 12yz + 10x + 17y + 5z = 481$$
In order to find x + y + z, we'll need to modify the equation and maybe solve for another variable.

If we group the terms involving x, we get:
$$(6yz)x + (30y + 18z + 10)x = 481 - (12yz + 17y + 5z)$$

We can now observe that the left side must be divisible by x, and the right side must be divisible by the expression in parentheses. However, no matter what values of y and z we choose, there is no guarantee that 481 - (12yz + 17y + 5z) will be divisible by the sum 30y + 18z + 10.

Thus, it seems that there is no solution with positive values for x, y, and z.