Solving Equation |x|+|y|+|z|=10 for I: A Comprehensive Guide

[Saitama]

Homework Statement

If in an equation |x|+|y|+|z|=10, x,y,z $\in$ I, then the number of solutions are
A)528
B)402
C)666
D)None

Homework Equations

The Attempt at a Solution

I am clueless on this one. It looks to me that it represents 8 planes but I don't think that's going to help me here.

 

[Coto]

All values of x,y,z are within [-10,10], in order for that equation to be satisfied. How many different ways can you add 3 integers (x,y,z s.t. 0 \leq x,y,z \leq 10) to be equal to 10. Don't worry about negatives for now.

 

[Saitama]

All values of x,y,z \leq 10, in order for that equation to be satisfied. How many different ways can you add 3 integers (x,y,z s.t. 0 \leq x,y,z \leq 10) to be equal to 10. Don't worry about negatives for now.

For 0<= x,y,z <=10, the number of solutions are 12C2. I tried this before but then got stuck when it came to my mind that I need to consider the negatives too.

 

[Coto]

How many solutions do you have when you have 0 \leq y,z \leq 10, and -10 \leq x \leq 0? I would iterate your process fixing one, then 2, then 3 variables to be negative. You could then setup an equation that adds each of those solution sets together and subtract duplicate solutions.

 

[haruspex]

Yes, it's 8 planes, specifically the surface of a regular octahedron. So you can use symmetry. This is the same as Coto suggests, but using a more geometric (visual) approach. E.g. count one plane, multiply by 8, then set about subtracting the double counts: each edge was counted twice, each vertex ... how many times?

 

[Saitama]

Yes, it's 8 planes, specifically the surface of a regular octahedron. So you can use symmetry. This is the same as Coto suggests, but using a more geometric (visual) approach. E.g. count one plane, multiply by 8, then set about subtracting the double counts: each edge was counted twice, each vertex ... how many times?

During the examination, I used the same approach but I did not end up with the right answer.

Considering the plane x+y+z=10, the integer solutions where x,y,z>0 are 9C2. Multiplying by 8 and adding up the edges, I get 9C2*8+10*6+1 (+1 for the origin)=349 which is incorrect. I don't see how this is wrong.

EDIT:
Considering the plane x+y+z=10, the integer solutions where x,y,z>0 are 9C2. We have 12 edges. The number of points on those edges are (not including the vertices) 9*12 and adding up the vertices i.e. 6
I get 9C2*8+12*9+6=402. Woops, looks like my approach was correct, did some silly mistakes in the exam. -_-

I looked at the given solution, it is given that required solutions=9C1*3C1*2^2+9C2*2^3+1*3C2*2=402. Can you explain me why did they write 3C1 and 2^2?

Thank you haruspex and Coto!

 

[haruspex]

Can you explain me why did they write 3C1 and 2^2?

This is for number of solutions along edges of the octahedron, excluding the vertices. The are 9 such along each edge, so you're asking why they write 3C1*2^2 for the number of edges. An edge corresponds to one of the variables being 0: 3C1. The remaining two variables are nonzero and can have either sign independently: 2^2.

 

[Saitama]

This is for number of solutions along edges of the octahedron, excluding the vertices. The are 9 such along each edge, so you're asking why they write 3C1*2^2 for the number of edges. An edge corresponds to one of the variables being 0: 3C1. The remaining two variables are nonzero and can have either sign independently: 2^2.

Got it, thank you haruspex!