If x, y, and z are positive integers such that 6xyz + 30xy + 18xz + 12yz + 10x + 17y + 5z =481, find x + y + z.
3 answers
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, $x+y+z = 1+1+1 = \boxed{3}$.
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.
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.
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.
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.