# How many solutions does the equation have?

1. Oct 3, 2015

### terryds

1. The problem statement, all variables and given/known data
If $0 \leq x_1 < 6, x_2 \geq 0, x_3 > 5$, how many solutions does $x_1+x_2+x_3=17$ have ?

A.46
B.57
C.68
D.79
E.89

2. Relevant equations

3. The attempt at a solution

For x3, $17 - x_1 - x_2 > 5$
For x2, $17 - x_1-x_3 \geq 0$
For x1, $0 \leq 17 - x_2 - x_3 < 6$

Then, I don't know what to do next..

2. Oct 3, 2015

### Staff: Mentor

Is this the whole problem?

I would expect them to say that x1, x2 and x3 are integers.

What can you say about x1? What values can it take?

Similarly for the others.

Basically this is a counting problem, ie for each x1 value count the number of x2 and x3 values that make the x1+x2+x3=17 true.

3. Oct 3, 2015

### terryds

So, there is no better way than trying each one ??
I think there's a better way since all of the options are a bit big numbers (46 is the smallest)

4. Oct 3, 2015

### Staff: Mentor

There probably is a better way but since you dont see it yet then why not try to count them.

Pick x2 and set it to 0 then how many choices are there for x1 and x3?

5. Oct 3, 2015

### terryds

There are seven choice..
(0,17),(1,16),...,(6,11)

6. Oct 3, 2015

### Staff: Mentor

Close but x1 =/= 6

now try x2=1 then x2=2 ...

If you can spot the pattern thats great but you might be able to eliminate some of the choices like say is 89 too high a count?

7. Oct 3, 2015

### terryds

So there is 6 solutions for x2 = 0
If x2 = 1, then there will be 6 solutions also
So, for x2 = 0 until x2 = 6, it will have 6 solutions.
But for x2=7, there are 5 solutions.
For x2 = 8, there are 4 solutions... and so on

So, by trying each integer for x2 value, there are 7*6+5+4+3+2+1 = 57 solutions.. Got it! Thankss