Coin Combinations: Solving Part a) and Questions for Part b)

[Workout]

Homework Statement

6. a) In your pocket you have some nickels, dimes, and quarters. There are 20 coins altogether and exactly twice as many dimes as nickels. The total value of the coins is $3.00.
Find the number of coins of each type.
b) Find all possible combinations of 20 coins (nickels, dimes, and quarters) that will make exactly $3.00.



I solved part a) and got 4 nickels, 8 dimes, 8 quarters.

I really don't know how to do (b). I could make a chart, but is there a mathematical procedure to doing this? I don't know how to put it in matrice form.

 

[Mark44]

Homework Statement

6. a) In your pocket you have some nickels, dimes, and quarters. There are 20 coins altogether and exactly twice as many dimes as nickels. The total value of the coins is $3.00.
Find the number of coins of each type.
b) Find all possible combinations of 20 coins (nickels, dimes, and quarters) that will make exactly $3.00.



I solved part a) and got 4 nickels, 8 dimes, 8 quarters.

I really don't know how to do (b). I could make a chart, but is there a mathematical procedure to doing this?

I don't believe there is, but it helps to think in a systematic way. For example, you can't have only nickels or only dimes, so that eliminates lots of possiblities. Also, you can't have only quarters, because that would leave you with too much money.

Start with the largest number of quarters for which you have a chance of getting $3, and then see if some combination of nickels and dimes gives you the right amount.

Work you way down with 1 less quarter each time, looking at the number of nickels and dimes that will work.

I don't know how to put it in matrice form.

The word is matrix. Its plural is matrices. There is no word "matrice" in English. You can thank the Romans for this contribution to the complexity of English.

 

[Mark44]

BTW, this is not a Calculus problem, so I'm moving it to the Precalc section.

 

[haruspex]

b) reduces to making a total of 60 from twenty 1s, 2s and 5s.

Let the number of 5s be n. If that's odd then there must also be an odd number of 1s, so it splits into two cases. E.g. for n=2k, there's 60-10k to be made from 20-2k 2s and 1s. If there are m 2s we have 2m+(20-2k-m)=60-10k. After simplifying, you can quickly see what combinations are possible.