Linear Algebra Money Question

  • #1
bleedblue1234
109
0

Homework Statement



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?

Homework Equations



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

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

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.
 

Answers and Replies

  • #2
E_M_C
43
0
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
azizlwl
1,065
10
You can narrow the selection.
$1 can only be in a group of 5.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
43,021
970
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 by a moderator:
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement



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?

Homework Equations



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

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

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.

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
HallsofIvy
Science Advisor
Homework Helper
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Oh, well- if you want to do it the easy way!
 
  • #7
bleedblue1234
109
0
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
E_M_C
43
0
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.

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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