Does other solutions exist for $29x+30y+31z = 366$?

Question

I was asked this trick question:

If $29x + 30y + 31z = 366$ then what is $x+y+z=?$

The answer is $12$ and it is said to be so because $29$ , $30$ and $31$ are respectively the number of days of months in a leap year. Therefore $x + y + z$ must be $12$, the total number of months.

How accurate is this? Is it possible to say so with just a single equation? Are there not other solutions to the equation? If yes, how can one proceed to find other solutions?

Answer 1

Not true. For instance, $x=y=0$ and $z=366/31$ is another solution, whose sum is not $12$.

Answer 2

With $x,y,z$ being reals, no, there's tons of answers. With $x,y,z$ be natural numbers, $12$ is the only answer. Proof: $11$ is too small, because even if all $11$ was in the biggest number, $11\cdot 31=341<366$, and $13$ is too big because $13\cdot 29=377>366$.

The "middle" case if you allow negative integers but not rationals/reals, I'm not sure about

Answer 3

Here is the solution to Alan's "middle case", where the three variables are allowed to be integers:

Choose any integer for $z$ and demand that $x=-y$ (so that $x+y=0$); then $x+y+z=z$, so if we can find a value for $y$ that satisfies the 366 constraint, we show that the $x+y+z$ can be any integer.

This is possible: observe that

$$29x+30y+31z ~=~ 29(x+y)+y+31z ~=~ y+31z,$$

so that if we choose $y=366-31z$, we get $(366-31z)+31z=366.$