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Linear Algebra Money Question

  1. Sep 14, 2012 #1
    1. The problem statement, all variables and given/known data

    I have 32 bills in my wallet in the denominations $1, $5, and $10, worth $100 in total. How many of each denomination do I have?

    2. Relevant equations

    A= # $1 bills
    B= # $5 bills
    C= # $10 bills

    A+B+C = 32
    1A+5B+10C = 100

    3. The attempt at a solution

    So I attempted to solve for C in terms of A and B in terms of A but I'm getting nowhere.
     
  2. jcsd
  3. Sep 14, 2012 #2
    Hi bleedblue1234,

    You can only solve for n variables when you have n linearly independent equations. In this case, you have 3 variables and 2 linearly independent equations, so you're one equation short.

    But if you choose a value of zero for A, B or C then you reduce the problem to 2 variables and 2 linearly independent equations. What do you get when you try out the different combinations?

    Be careful: There is more than one solution.
     
  4. Sep 14, 2012 #3
    You can narrow the selection.
    $1 can only be in a group of 5.
     
  5. Sep 14, 2012 #4

    HallsofIvy

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    This not, strictly speaking, a "linear algebra" problem, but a "Diophantine equation" because the "number of bills" of each denomination must be integer. Letting "O", "F", and "T" be, respectively, the number of "ones", "fives" and "tens", we must have O+ F+ T= 32 and O+ 5F+ 10T= 100. Subtracting the first equation from the second, 4F+ 9T= 68.
    Now you can use the standard "Eucidean algorithm" to find all possible integer values for F and T and then find O.
     
    Last edited: Sep 14, 2012
  6. Sep 14, 2012 #5

    Ray Vickson

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    You can solve for A and B in terms of C, just by solving the two simple equations
    A + B = 32 - C
    A + 5C = 100 - 10C.

    Now you can plug in C = 0, 1, 2, ... and see which values (if any) give you non-negative integer values of A and B.

    RGV
     
  7. Sep 14, 2012 #6

    HallsofIvy

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    Oh, well- if you want to do it the easy way!
     
  8. Sep 14, 2012 #7
    Ya I just set the equations equal and restricted B and C to be natural numbers and just checked which B would give me the correct C, which in tern gave me the correct A. Thank you.
     
  9. Sep 14, 2012 #8
    Be careful with that. Most people (in my experience) define the natural numbers as N = {1, 2, 3, ...} which doesn't include zero. So if you're restricting B and C to natural numbers, as defined above, you may be cheating yourself out of a solution. As I mentioned earlier, there is more than one solution.
     
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