Question

Need help with a discrete math story problem.

Answer 1

@phi

Answer 2

any ideas ?

Answer 3

Well I would need to write some kind of equation.

Answer 4

A:

if B = bills, D = dimes, q = quarters, n = nickles

I know that

B = 2D + 1Q
N = Q

Answer 5

For machine A

Answer 6

Machine B:

Q = B + 2N

D = B

Answer 7

I would try to find what I have after using A then B
then do it a few more times, and hope I can see a pattern...

Answer 8

Also, I have some notes which I don't really understand.

Answer 9

Ok, I tried that.

After one trip to machine A changing a dollar bill:

I get 2D + 1Q

I get 2B and B + 2N in machine B.


So altogether I have 3B + 2N

Answer 10

they almost ring a bell with me. But I would have to look this stuff up... I don't know it off the top of my head.

Answer 11

So I know that after one trip to both the machines, 1B becomes 3B + 2N

and if I go back to the machine I assume 3B would become 9B + 3N and the 2N would become 2Q which would become 2B + 4N

So I would be left with 11B + 7N after the second time.

Answer 12

I just don't understand how to find the equation to calculate the amount each time we loop.

Answer 13

I get 11B + 10N the 2nd time through

Answer 14

oh woops, I changed the B into Ds and Ns instead of Ds and Qs

Answer 15

So in the notes,

Nn+1 = 2(Bn + Nn)
Bn+1 = 3Bn + Nn

These equations work when I plug the numbers in to give us the next B and N

Answer 16

yes. and your notes find a "characteristic equation" with roots 1 and 4
That should be helpful, but I would have to study up on this stuff...

Answer 17

So Nn+2 = 5Nn+1 - 4Nn N > 0

is the characteristic function?

Answer 18

yes. See http://www.wikihow.com/Solve-Recurrence-Relations
method three for how I think you solve this.

Answer 19

oops, method 4

Answer 20

Hmm, this looks complicated. :/

Answer 21

Let me look at a few minutes. Meanwhile, read over Method 4: Linear

Answer 22

ok

Answer 23

In your problem, your characteristic equation for the number of nickels N
is
\[ N_{n+2} = 5N_{n+1} - 4N_n \\
x^2 - 5x + 4 = 0 \]
which factors
\[ (x-1)(x-4)=0 \\
x=1, \ x=4 \]

the roots are 1 and 4
according to method 4 (why this works, I don't know)
\[ N_n= c_1 1^n + c_2 4^n \]
where 1 and 4 are the roots

1^n is always 1 so this is the same as
\[ N_n= c_1 + c_2 4^n \]

Answer 24

now we need to find the constants.
when n=1 we found 3B + 2N , so N = 2
when n=2 we found 11B+10N, so N=10 at n=2
use those numbers to find c1 and c2

Answer 25

Do I need to use step 6 for this?

Answer 26

or am I solving for the equation above?

Answer 27

same idea. but for your problem we use your numbers.

Answer 28

in other words,
\[ c_1 + c_2 4^n=N_n\\
n=1, \ c_1 + c_2 4^1= 2 \\
n=2,\ c_1 + c_2 4^2= 10\]

Answer 29

c1 = 2-c2(4)

c2(8) + 2 - c2(4) = 10

c2(32) = 8

c2 = 1/4

Answer 30

c1 = 10 - c2(8)
c1 = 10 - 2
c1 = 8

Answer 31

Would that be right?

Answer 32

I think 4^2 = 16

Answer 33

oops

Answer 34

I would just subtract the 2 equations

Answer 35

Not sure how to do that

Answer 36

c1 = 2 - c2(4)

2 - c2(4) + c2(16) = 10


2 - c2(20) = 10

c2 = -8/20

Answer 37

you can use substitution, also.

but for subtraction:
\[ c_1 + 16 c_2 = 10\\
- c_1 -4 c_2 = -2\\
----------\\
12 c_2=8\\
c_2= \frac{8}{12}=\frac{2}{3}
\]

Answer 38

ah nice, that is much quicker, lol. Guess I still didn't get the right answer.

Answer 39

ah ok, I see the mistake.

Answer 40

you added 4 instead of subtracting 4 in your work. you should get
2- 12 c2 = 10

Answer 41

c1 = 14/3

Answer 42

your algebra is unacceptable

Answer 43

c1 = 2 - c2(4)
c1 = 2 - ⅔ *4= 2 - 8/3 = 6/3 - 8/3 = -2/3

Answer 44

I made another dumb mistake, sorry.

Answer 45

Multiplied by the wrong number in the first step.

Answer 46

so the number of nickels is at step n is
\[ N_n = \frac{2(4^n -1)}{3} \]
the number of bills B is 1 more than the number of N, so we can say
\[ B_n= \frac{2(4^n -1)}{3}+1= \\\frac{2\cdot 4^n -2}{3}+\frac{3}{3}= \\\frac{2\cdot 4^n +1}{3} \]

Answer 47

Awesome, thank you so much for the help.

Answer 48

now when you learn why, explain how this works, because it is a mystery to me

Answer 49

haha, if I ever learn how I will let you know. :D

Answer 50

This was the last problem on my second to last homework in the class though, since class is almost finished, I doubt I will ever figure out exactly how this works. :/

Answer 51

But thanks again! I will try to be less dull when it comes to algebra in the future. -_-