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?

If any values will do, then HowDoIMath's answer is a counterexample, and a guide to others. Just pick numbers for x and y, and solve for z — Alan, Aug 08 '15 at 21:06
@Alan Ok. But Will the result change if it is limited to Integer values? — Aswin P J, Aug 08 '15 at 21:07
In this case you know one solution, because february has 29 days (x=1, all four years) and you have seven months with 31 days and 4 months with 30 days. — callculus42, Aug 08 '15 at 21:07
Answered below with the implicit assumption that they are natural numbers, with a proof — Alan, Aug 08 '15 at 21:11
Another answer is $x=2,y=2,z=8$, though Alan has shown it has the same x+y+z — Empy2, Aug 08 '15 at 21:27
If you like one of the answers best you should tag it as best answer, but if you are totally split, you should choose HowDoIMath since he has the least rep ;) — Eric Stucky, Aug 09 '15 at 01:06

score 7 · Answer 1 · answered Aug 08 '15 at 21:11

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

score 5 · Accepted Answer · answered Aug 08 '15 at 21:05

5

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

answered Aug 08 '15 at 21:05

Mankind

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Thank you for the response Sir. Didnt think of that :D Curious as to how the result might change if the values of x,y and z are limited to integers. Will the result change? In that case is x+y+z always equal to 12? Kindly update your answer. – Aswin P J Aug 08 '15 at 21:08
No problem, @Hope . I think the integer case has been covered in the other answers. – Mankind Aug 08 '15 at 21:23

score 4 · Answer 3 · answered Aug 08 '15 at 21:20

4

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 $(366-31z)+31z=366.$

answered Aug 08 '15 at 21:20

Eric Stucky

12,758
3
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Bonus question: Describe all numerical monoids for which the solutions to $29x+30y+31z$ take on a finite set of values. (I have no idea if this is reasonable; seems like not.) – Eric Stucky Aug 08 '15 at 21:23

score 0 · Answer 4 · answered Jul 12 '16 at 12:28

0

if assume x,y,z are integers, then x=1,y=4,z=7 or x=2,y=2,z=8 or x=3,y=0,z=9

answered Jul 12 '16 at 12:28

Wayne Zhang

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

4 Answers4