a money puzzle question (equations) (1 Viewer)

[modular]

ummm if someone could help me with this id be very greatful, i started off doin a bunch of linear equations with 2 variables but then got all confused -_-

"Imagine you wanted to change an Australian 10$ note into coins of denomination 20 cents or higher. In how many different ways could you do this using only coins of the same denomination? Using more than one coin of denomination?"

i was doing equations, and trial of error, but then i was just like BLAH

 

[foram]

ummm if someone could help me with this id be very greatful, i started off doin a bunch of linear equations with 2 variables but then got all confused -_-

"Imagine you wanted to change an Australian 10$ note into coins of denomination 20 cents or higher. In how many different ways could you do this using only coins of the same denomination? Using more than one coin of denomination?"

i was doing equations, and trial of error, but then i was just like BLAH

I think this is more of a permutations and combinations question.

 

[lyounamu]

I think this is more of a permutations and combinations question.

I reckon. I got the first question but the 2nd question is like WTH? I don't understand this.

EDIT: 2nd question has got TOO many possibilities.

 

[modular]

i think i worked it out

the first part is just, for eg, 20 cents
20x = 1000
so 50x 20 cent coins will give 10 dollars

then for the second part, u do it with all the combos

for eg, for the combination of 20cent donimations and 50c coins

wed go 20x + 50y = 1000
x = 1000 - 50y / 20

so the range must be between 1-20 to give x a positive whole number, and obviously must be in successions of 2 because we need x to be a whole number (2, 4 into 20 has no remainder etc etc)

so then u draw up table, plug in values 2,4,6, 8 etc all the way to 18 (because 20 is with no 20cents)

and wallah, the table is ur possible combo for using 20 cents and 50 cents to change a 10$ dollar note

WOW. fuck doing that for every other possible combination, ie 50 cents and 1 dollar, 50 cents and 2 dollars, 20 cents and 1 dollar, 20 cents and 2 dollars... then need to go on to triple variables... 1 dollar, 20 cents, 50 cents etc etc

way to much effort

 

[webby234]

ummm if someone could help me with this id be very greatful, i started off doin a bunch of linear equations with 2 variables but then got all confused -_-

"Imagine you wanted to change an Australian 10$ note into coins of denomination 20 cents or higher. In how many different ways could you do this using only coins of the same denomination? Using more than one coin of denomination?"

i was doing equations, and trial of error, but then i was just like BLAH

I may have solved it
Ok my equation is 2x₁ + 5x₂ + 10x₃ + 20x₄ = 100

You can use the fact that there are limited numbers of possible ways of making 100 when you're restricted to a multiple of 10 plus a multiple of 20.

So i'm temporarily combining x₃ and x₄

So for example you can have:
20 + 0 + 80
20 + 10 + 70
20 + 20 + 60 etc. down to
20 + 80 + 0

If you are making up 80 with sums of multiples of 5 and 2, you can have 9 different ways, because for every even multiple of 5 less than or equal to 80, there is a corresponding multiple of 2 that adds up to 80. ie in this case x₃ can be 0, 2, 4, 6, 8, 10, 12, 14 or 16. So if you have one twenty cent coin, you have 9 (for 20 + 0 + 80) + 8 (for 20 + 10 + 70) + 7 + 6 + 5 + 4 + 3 + 2 + 1(for 20 + 80 + 0) = 4*10 + 5 = 45 different solutions.
Similarly, for two twenty cent coins, you would have 7 + 6 + 5 +...+1 different solutions. etc.

So total you have 11 + 10 + 2(9 + 8) + 3(7 + 6) + 4 (5 + 4) + 5 (3 + 2) + 6 different solutions? Does anyone follow my logic or am i not making sense?

 

[modular]

rofl i dont know, but u confused the FUCK out of me

 

[Lolita25]

Ok my equation is 2x1 + 5x2 + 10x3 + 20x4 = 100

but i see someone already beat me to the eqn....so GOOD LUCK!

 

[modular]

way to confused, need sleep


hope i can work out how to do it, its for an assignment

 

[webby234]

Lol it's difficult to explain - it makes sense to me .

Basically I'm fixing the number of 2 dollar and 1 dollar coins and then working out how many possibilities of 20 and 50 cent coins there are. Do you understand why my original equation is as it is?

So lets say you have 2 two dollar coins and 2 one dollar coins.
By my equation, you have:
40 (twenty times the number of two dollar coins) + 20 (for the one dollar coins) + 5 times the number of 50 cent coins + 2 times the number of 20 cent coins = 10

So you have 40 + 20 + 5x₂ + 2x₁ = 100
5x₂ + 2x₁ = 40

x2 and x1 have to be integers, and as 2x₁ is even, then 5x₂ must be even (because they add to give an even number). So the possible ways are to have 0 + 40, 10 + 30, 20 + 20 and 30 + 10, 40 + 0. So for 2 two dollar and 2 one dollar coins, you have five different ways. Similarly, if you had 2 two dollar coins and only 1 one dollar coin, you would be looking for ways to have 5x₂ + 2x₁ = 50, and by the above argument there are 6 such ways. In general, there are n + 1 different solutions to 5x₂ + 2x₁ = 10n.

If you have no twenty cent coins, then you can have anywhere from 0 to 10 ten cent coins, so n can range from 0 to 10 and n + 1 varies from 1 to 11. So you have 11 + 10 + 9 +...+ 1 solutions for no twenty cent coins.

If you have one twenty cent coin, the rest can only add up to 80, so n can only be between 0 and 8, so you have 9 + 8 + 7 +...+ 1 solutions for one twenty cent coin... and so on down to 5 twenty cent coins (for which there's only one solution).

Lol it seemed so much simpler in my head . I don't actually know whether that will make more sense to you. It's difficult for 2U though - there should be a simpler way?

 

[lolokay]

I got 161. Assuming only denominations are 20c, 50c, $1, $2,

20c will always come in groups of 5, and 50c in groups of 2, ie $1 groups

there are 6 ways to choose the number of $2 coins,

The number of combinations of other coins, using combinations with repetition ([n+k-1]!/k![n-1]!) forumula for each of 0,1,2,3,4,5 $2 coins is, respectively: 66, 45, 28, 15, 6, 1.

So total number = 161

 

[webby234]

Yeah my solution works out to the same number, but I was trying to just use 2 unit stuff .