Linear Algebra Money Question

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 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. You can narrow the selection.$1 can only be in a group of 5.

HallsofIvy
Homework Helper
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.

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Ray Vickson
Homework Helper
Dearly Missed

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

HallsofIvy