How many solutions does the equation have?

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The equation x1 + x2 + x3 = 17 has constraints on the variables: 0 ≤ x1 < 6, x2 ≥ 0, and x3 > 5. The problem is approached as a counting problem, where for each integer value of x1, the corresponding values of x2 and x3 are calculated to satisfy the equation. By systematically testing values for x2, it is determined that there are 57 valid combinations of x1, x2, and x3 that meet the criteria. The solution involves recognizing patterns in the counts based on the chosen values for x2. Ultimately, the total number of solutions is confirmed to be 57.
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Homework Statement


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

Homework Equations

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..
Please help
 
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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.
 
jedishrfu said:
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.

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)
 
There probably is a better way but since you don't 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?
 
jedishrfu said:
There probably is a better way but since you don't 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?
There are seven choice..
(0,17),(1,16),...,(6,11)
 
Close but x1 =/= 6

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

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