Total possible solutions to the equation

[dancergirlie]

Homework Statement

Find the number of solutions to the equation
2x + 3y + 6z = 73
where x, y, and z are non-negative integers.

Homework Equations

The Attempt at a Solution

I don't know what to do here, usually I'd like to make a binary sequence to find out the number of possibilities, but this time there are restrictions (because of the 2,3, and 6) so I don't know what to do? Am I supposed to make a generating function or something??

All I know is that x can be all even numbers from 0 to 70
y can be all multiples of 3 from 0 to 69
and z can be all multiple of 6 from 0 to 66


Any help would be GREAT!

 

[Mark44]

Homework Statement

Find the number of solutions to the equation
2x + 3y + 6z = 73
where x, y, and z are non-negative integers.

Homework Equations

The Attempt at a Solution

I don't know what to do here, usually I'd like to make a binary sequence to find out the number of possibilities, but this time there are restrictions (because of the 2,3, and 6) so I don't know what to do? Am I supposed to make a generating function or something??

All I know is that x can be all even numbers from 0 to 70
y can be all multiples of 3 from 0 to 69
and z can be all multiple of 6 from 0 to 66

No, 2x is an even integer from 0 through 70, 3y is a multiple of 3 from 0 through 72, and 6z is a multiple of 6 from 0 through 72. Since 2x and 6z will always be even, 3y must be odd, which eliminates even possibilities for y.

I think that what I would do is look at the possible values for 6z, which are 0, 6, 12, 18, ..., 72. For each one I would look at the combinations of the other two variables that add to 73. It's been a long time since I studied number theory, and it was just one class, so there might be a better technique that's applicable that I don't remember.

 

[dancergirlie]

Yes, but when you look at the equation if y was a multiple of 3 from 0 through 72 then what variable would equal one to make the equation equal 73 since there is no number that could equal one... and if z was a multiple through 72 then what one would equal one once again? I considered looking at the combinations but i think there would be hundreds of them, so I thought maybe there was an easier way... i guess I could look at it :/

 

[Mark44]

Yes, but when you look at the equation if y was a multiple of 3 from 0 through 72 then what variable would equal one to make the equation equal 73 since there is no number that could equal one

I understand part of what you're saying, but some of the things you're saying don't make any sense. E.g., "make the equation equal 73" - an equation isn't equal to anything; an equation says that two expressions have the same value; "there is no number that could equal one" - well, 1 = 1, for starters.

I didn't say anything about y being a multiple of 3. All I said was that y had to be odd. You seem to be confusing 2x and x, 3y and y, and 6z and z.

The possible values for 6z are 0, 6, 12, 18, 24, ..., up to 72, which means that z could be equal to 0, 1, 2, ... , up to 12. As you pointed out, z=12 is not a possible solution, since there are no values for x or y that produce 1. So the possible values for z are 0, 1, 2, ..., 11. Can any of these values be eliminated as well?
The possible values for 3y are 0, 3, 6, 9, 12, 15, ..., up to 72, which means that y could be 0, 1, 2, ..., up to 24. For the reason I already mentioned, y can't be an even value, so y has to be one of 1, 3, 5, ..., 23. Note that some of these values are multiples of 3, but a lot of them aren't.

 

[dancergirlie]

Is there a simpler way to write this rather than showing all the cases? Because for z=0, there are going to be several possible solutions. Is there a way I can figure this out using combinations of some sort or is listing the only way that it can be done?

 

[dancergirlie]

wait, nevermind, i thought I had to go through all the values of x, however I can show that for y=1,3,5,9,11,13,15,17,19,21,23 that the equality will work... That isn't too bad to list... Also then for z=1, 6z=6, then the possible y values would be y=1,3,5,9,11,13,15,17,19,21. And then for 6z=12, the y values would be y=1,3,5,9,11,13,15,17,19. And as z increases by 6, then one value of the y is dropped. Am I right?

 

[Mark44]

wait, nevermind, i thought I had to go through all the values of x, however I can show that for y=1,3,5,9,11,13,15,17,19,21,23 that the equality will work... That isn't too bad to list... Also then for z=1, 6z=6, then the possible y values would be y=1,3,5,9,11,13,15,17,19,21. And then for 6z=12, the y values would be y=1,3,5,9,11,13,15,17,19. And as z increases by 6, then one value of the y is dropped. Am I right?

Sounds like you're on the right track. Again, there might be a more elegant way, but this should work. The more possibilities that you can eliminate, the fewer remaining possibilities you'll need to check. For example, if z = 11 (hence 6z = 66), the other two numbers have to add up to 7, which can happen for only a couple of values of x and y.

 

[dancergirlie]

yeah, I added it all up and I got 78 total possibilities. Thanks for the help!