A random maths puzzle

[Momo.05]

Laura is in charge of lighting the rock palace for upcoming willy nilly wild ones concert. Each light fixture supplies exactly 1,000 watts of power to light the bulbs in the fixture. Laura can use any combination of 150 watt, 100 watt, 75 watt, or 60 watt bulbs, but the total number of watts must be 1000. How many diffrent combinations of bulbs could laura use in a light fixture?

can someone help me solve this? Please. Thanks.

[TrueTears]

We need to find how many distinct there are.

Let be the number of nonnegative ordered 4-tuple that solve

We need to find

Define







Now

Thus and so on.

Our aim is the find the coefficient of the term, which looks like a recurrence relation.

[QuantumJG]

We need to find how many distinct there are.

Let be the number of nonnegative ordered 4-tuple that solve

We need to find

Define







Now

Thus and so on.

Our aim is the find the coefficient of the term, which looks like a recurrence relation.

I'm confused with that A(x),B(x)... notation. :S

[TrueTears]

It is generating functions.

[QuantumJG]

So,

- A(x) = x⁰⁽³⁰⁾ + x¹⁽³⁰⁾ + x²⁽³⁰⁾ + x³⁽³⁰⁾ ...

It's generating functions but I don't know how these relate to the problem.   

[kamil9876]

Hi. Let a,b,c,d be the number of 150,100,75,60 wat bulbs used respsectively.

We have:

150a+100b+75c+60d=1000

And now we will keep reducing this equation:

30a+20b+15c+12d=200

From here on we see that c must be even and d must be a multiple of 5, hence let c=2k, d=5x:

30a+20b+30k+60x=200

3a+2b+3k+6x=20

Now we can see that b cannot be a multiple of 3, hence b is of the form b=3m+1 or b=3m+2 (mutually exclusive cases). By plugging in the second case we get:

3a+2(3m+2)+3k+6x=20
3a+6m+3k+6x=16
3(a+2m+k+2x)=16 which implies 16 has 3 as a factor, a contradiction. Hence only the first case is possible:

3a+2(3m+1)+3k+6x=20
3a+6m+3k+6x=18
a+2n+k+2x=6

And now find how many different solutions there are to that equation. Which is done manually or by computer or maybe generating functions (dunno what is quickest so far).

[TrueTears]

So,

- A(x) = x⁰⁽³⁰⁾ + x¹⁽³⁰⁾ + x²⁽³⁰⁾ + x³⁽³⁰⁾ ...

It's generating functions but I don't know how these relate to the problem.   

Ahmad taught me alot about generating functions (eg, http://vcenotes.com/forum/index.php/topic,19896.msg210235.html#msg210235), you could also have a read in AnC's page 136

[TrueTears]

Hi. Let a,b,c,d be the number of 150,100,75,60 wat bulbs used respsectively.

We have:

150a+100b+75c+60d=1000

And now we will keep reducing this equation:

30a+20b+3c+12d=200

From here on we see that c must be even, hence let c=2k:

30a+20b+6k+12d=200

15a+10b+2k+6d=100

From here on, we see that a must be even, hence let a=2m:

30m+10b+2k+6d=100

15m+5b+k+3d=50

So we wanna know the numbers of non-negative integer solutions to that equation.

This sort of simplifies the problem to smaller numbers, but doesn't simplify it as much as I hoped. To proceed further you can use a similair approach as TT did with the generating functions.

let be the number of ways of solving the problem for instead of 50. We want to know .





Now you just multiply it out and equate coefficients etc. and you get some more equations that solve it. I'm too tired to proceed further but I don't know how much it simplifies the problem too, I have the pessimistic feeling that it can't probably be simplified to the point that we  don't  need to do a fairly large amount of tedious listing.

wait how did you get 30a+20b+3c+12d=200?

[kamil9876]

ahh yes sorry, 75/5=15. lol ill edit now

[TrueTears]

ahh yes sorry, 75/5=15. lol ill edit now

lol yeah, the other steps follow on.

[kamil9876]

ahh yes sorry, 75/5=15. lol ill edit now

wow, noticing that blunder led to a better simplification, check out new edited post

[QuantumJG]

ahh yes sorry, 75/5=15. lol ill edit now

wow, noticing that blunder led to a better simplification, check out new edited post

Do we learn how to do this at uni?

[TrueTears]

ahh yes sorry, 75/5=15. lol ill edit now

wow, noticing that blunder led to a better simplification, check out new edited post

wow much better, yeah now generating functions work better...

[TrueTears]

ahh yes sorry, 75/5=15. lol ill edit now

wow, noticing that blunder led to a better simplification, check out new edited post

Do we learn how to do this at uni?

You can have a read of H.S.Wilf, Generating Functionalogy.

[QuantumJG]

ahh yes sorry, 75/5=15. lol ill edit now

wow, noticing that blunder led to a better simplification, check out new edited post

wow much better, yeah now generating functions work better...

Could someone tell me the basics of how you generate functions?