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# Department of Science, Math, & Technology

E = mcÂ˛
99 cents
Posted: Posted April 9th by Xhin
 There are various ways of making 99 cents from change, all of which require at least 4 pennies: 9 dimes, 1 nickel, 4 pennies 1 quarter, 3 dimes, 5 nickels, 44 pennies 3 quarters, 2 dimes, 4 pennies That last one is how you make it in the least amount of coins possible. This is because of the following four rules: 1. A quarter counts as one coin 2. A dime counts as one coin 3. A nickel counts as one coin 4. A penny counts as one coin However, if a quarter counted as three coins, then suddenly 9 dimes, 1 nickel, 4 pennies is going to be a better option. Can you find a set of these four rules where most or all of the ways of making 99 cents are equal?
There are 7 Replies

The first thing we'd have to do is list out all of the possible combinations of those four coins (quarter, dime, nickel, and penny -- so no half-dollars or mythical two-cent pieces or whatever) that could make 99c. For example:
*3Q, 2D, 4P
*1Q, 3D, 5N, 19P
*9D, 1N, 4P
*etc, etc, etc.

Then we'd have to treat Q, D, P, and N like variables and find values for said variables that can satisfy the conditions. My question to you, Xhin, is do we limit the combinations to your list above or use every single possible combination of coins?

In any case, you can solve this quite easily by making the variables equal to the number of pennies each coin would comprise: Q=25, D=10, N=5, P=1.

Or am I (as per the usual on these math questions) missing something huge?

Edited April 10th by Black Yoshi

1 quarter, 3 dimes, 5 nickels, 44 pennies

This does not equal 99 cents

Posted April 10th by Brandy

True...it's actually \$1.24. @Xhin:

Posted April 10th by Black Yoshi

Oops, I meant 0 nickels.

Posted April 10th by Xhin
Xhin
Nature is beautiful

My question to you, Xhin, is do we limit the combinations to your list above or use every single possible combination of coins?

Every single possible combination.

In any case, you can solve this quite easily by making the variables equal to the number of pennies each coin would comprise: Q=25, D=10, N=5, P=1.

That's the answer I got as well. I took it a step further though:

• Divide each "coin value" by its "coin count". This gives you a "net value".

• No matter what value system you use to assign "net values" to coins, the best strategy is always to use whatever has the most net value first.

So, in the normal case (where the coins are each equal to one coin), your net values are Q = 25, D = 10, N = 5, P = 1. This means you should use as many quarters as possible, then use as many dimes as possible, etc.

In your case, all the net values are equal to 1 so it doesn't matter what coins you use.

Now let's take some novel case:

• 1. Quarters = 5 coins
• 2. Dimes = 1 coin
• 3. Nickels = 1 coin
• 4. Pennies = 2 coins

Here, we calculate the net value of everything:

• 1. Quarters = 25/5 = 5nv
• 2. Dimes = 10/1 = 10nv
• 3. Nickels = 5/1 = 5nv
• 4. Pennies = 2/1 = 2nv

This is similar to the normal case, except that quarters and nickels are identical, and both are less efficient than dimes. So you're best off with

• 9 dimes, 1 nickel, 4 pennies (10nv -> 5nv -> 2nv)

• Posted April 11th by Xhin
Xhin
Nature is beautiful

Edited April 11th by Fox Forever

Make everything worth 0 coins and all combinations will be equal.

Posted Saturday by EN