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How many solutions does the equation have?

  1. Oct 3, 2015 #1
    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..
    Please help
     
  2. jcsd
  3. Oct 3, 2015 #2

    jedishrfu

    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.
     
  4. Oct 3, 2015 #3
    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)
     
  5. Oct 3, 2015 #4

    jedishrfu

    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?
     
  6. Oct 3, 2015 #5
    There are seven choice..
    (0,17),(1,16),...,(6,11)
     
  7. Oct 3, 2015 #6

    jedishrfu

    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?
     
  8. Oct 3, 2015 #7
    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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