# Linear Algebra Money Question

1. Sep 14, 2012

### bleedblue1234

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. Sep 14, 2012 ### E_M_C 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. 3. Sep 14, 2012 ### azizlwl You can narrow the selection.$1 can only be in a group of 5.

4. Sep 14, 2012

### HallsofIvy

Staff Emeritus
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
5. Sep 14, 2012

### Ray Vickson

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

6. Sep 14, 2012

### HallsofIvy

Staff Emeritus
Oh, well- if you want to do it the easy way!

7. Sep 14, 2012

### bleedblue1234

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.

8. Sep 14, 2012

### E_M_C

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.