You have ten U.S. coins that add up exactly to \$1.00

• Sep 4th 2009, 09:29 PM
TheBerkeleyBoss
You have ten U.S. coins that add up exactly to \$1.00
Hi, my name is Andy. I am currently taking linear algebra. My teacher assigned us a problem but I've come to an impasse.

(4 points) You have ten U,S. coins in current circulation that add up to exactly \$1.00. Find the solution that has the most types of coins. (Current circulation denominations: \$1.00, 50 cents, 25 cents, 10 cents, 5 cents, 1 cent.)

(variables are represented by the first letter of their coin name)

$50F + 25Q + 10D + 5N + P = 10$

$0.50F + 0.25Q + 0.10D + 0.05N + 0.01P = 1.00$

If you multiply the first equation by 10, and the second equation by 100, you can then set both equations equal to each other.

$500F + 250Q + 100D + 50N + 10P = 100$

$50F + 25Q + 10D + 5N + P = 100$

However, I'm not quite sure what to do from here. Isolate a variable? This worked with 3 variables (chicken, hen, rooster problem), but 5 variables is too much to do this with! or is it? RREF form? hmmmm....

The intended way to solve this was to isolate the variable with the highest value, F. I'm lost here.

Can anyone give me the next step? I'm not looking for a total answer, but just a lead in the right direction.
• Sep 5th 2009, 12:31 AM
CaptainBlack
Quote:

Originally Posted by TheBerkeleyBoss
Hi, my name is Andy. I am currently taking linear algebra. My teacher assigned us a problem but I've come to an impasse.

(4 points) You have ten U,S. coins in current circulation that add up to exactly \$1.00. Find the solution that has the most types of coins. (Current circulation denominations: \$1.00, 50 cents, 25 cents, 10 cents, 5 cents, 1 cent.)

(variables are represented by the first letter of their coin name)

$50F + 25Q + 10D + 5N + P = 10$

$0.50F + 0.25Q + 0.10D + 0.05N + 0.01P = 1.00$

If you multiply the first equation by 10, and the second equation by 100, you can then set both equations equal to each other.

$500F + 250Q + 100D + 50N + 10P = 100$

$50F + 25Q + 10D + 5N + P = 100$

However, I'm not quite sure what to do from here. Isolate a variable? This worked with 3 variables (chicken, hen, rooster problem), but 5 variables is too much to do this with! or is it? RREF form? hmmmm....

The intended way to solve this was to isolate the variable with the highest value, F. I'm lost here.

Can anyone give me the next step? I'm not looking for a total answer, but just a lead in the right direction.

Your first equation is wrong, you have 10 coins so:

$F + Q + D + N + P = 10$

CB
• Sep 5th 2009, 12:42 AM
CaptainBlack
Quote:

Originally Posted by TheBerkeleyBoss
Hi, my name is Andy. I am currently taking linear algebra. My teacher assigned us a problem but I've come to an impasse.

(4 points) You have ten U,S. coins in current circulation that add up to exactly \$1.00. Find the solution that has the most types of coins. (Current circulation denominations: \$1.00, 50 cents, 25 cents, 10 cents, 5 cents, 1 cent.)

(variables are represented by the first letter of their coin name)

$50F + 25Q + 10D + 5N + P = 10$

$0.50F + 0.25Q + 0.10D + 0.05N + 0.01P = 1.00$

If you multiply the first equation by 10, and the second equation by 100, you can then set both equations equal to each other.

$500F + 250Q + 100D + 50N + 10P = 100$

$50F + 25Q + 10D + 5N + P = 100$

However, I'm not quite sure what to do from here. Isolate a variable? This worked with 3 variables (chicken, hen, rooster problem), but 5 variables is too much to do this with! or is it? RREF form? hmmmm....

The intended way to solve this was to isolate the variable with the highest value, F. I'm lost here.

Can anyone give me the next step? I'm not looking for a total answer, but just a lead in the right direction.

That's not the way that I would do it, but if you do you may observe you cannot have any \$1 coins in the solution.

Then the next highest is F, and you can have 1 or 0 in a solution, assume its 1 and see where that leads, then assume its 0 and see where that leads.

(This does not look much like linear algebra to me, especially since the solution can be found virtually by inspection)

CB
• Sep 5th 2009, 01:44 AM
ANDS!
That's actually a similar problem that is in my linear algebra book. Introductory chapter sure, but it is reinforcing old ideas.
• Sep 5th 2009, 02:28 AM
TheBerkeleyBoss
This is the part where I get stuck though....once I set F equal to one. I can do this in my head without using linear algebra. However, my teacher did an example on the board (similar to this) putting the equations in a matrix and in reduced row echelon form. Thank you for all your responses so far.