How many different combinations of pennies, nickels, dimes and quarters use 48 coins to total $\$ 1.00$ ?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8
Solution 1

We want to combine 48 coins to get 100 cents.

Since the combined value of the coins is a multiple of 5 , as is the value of a combination of nickels, dimes and quarters, then the value of the pennies must also be a multiple of 5 .

Therefore, the possible numbers of pennies are 5, 10, 15, 20, 25, 30, 35, 40 .

We can also see that because there are 48 coins in total, it is not possible to have anything other than 35, 40 or 45 pennies. (For example, if we had 30 pennies, we would have 18 other coins which are worth at least 5 cents each, so we would have at least $30+5 \times 18=120$ cents in total, which is not possible. We can make a similar argument for 5, 10, 15, 20 and 25 pennies.)

It is also not possible to have 3 or 4 quarters. If we did have 3 or 4 quarters, then the remaining 45 or 44 coins would give us a total value of at least 44 cents, so the total value would be greater than 100 cents. Therefore, we only need to consider 0,1 or 2 quarters.

Possibility 1: 2 quarters

If we have 2 quarters, this means we have 46 coins with a value of 50 cents.

The only possibility for these coins is 45 pennies and 1 nickel.

## Possibility 2: 1 quarter

If we have 1 quarter, this means we have 47 coins with a value of 75 cents.

The only possibility for these coins is 40 pennies and 7 nickels.

Possibility 3: 0 quarters

If we have 0 quarters, this means we have 48 coins with a value of 100 cents.

If we had 35 pennies, we would have to have 13 nickels.

If we had 40 pennies, we would have to have 4 dimes and 4 nickels.

It is not possible to have 45 pennies.

Therefore, there are 4 possible combinations.

## Solution 2

We want to use 48 coins to total 100 cents.

Let us focus on the number of pennies.

Since any combination of nickels, dimes and quarters always is worth a number of cents which is divisible by 5 , then the number of pennies in each combination must be divisible by 5 , since the total value of each combination is 100 cents, which is divisible by 5 .

Could there be 5 pennies? If so, then the remaining 43 coins are worth 95 cents. But each of the remaining coins is worth at least 5 cents, so these 43 coins are worth at least $5 \times 43=215$ cents, which is impossible. So there cannot be 5 pennies.

Could there be 10 pennies? If so, then the remaining 38 coins are worth 90 cents. But each of the remaining coins is worth at least 5 cents, so these 38 coins are worth at least $5 \times 38=190$ cents, which is impossible. So there cannot be 10 pennies.

We can continue in this way to show that there cannot be $15,20,25$, or 30 pennies.

Therefore, there could only be 35,40 or 45 pennies.

If there are 35 pennies, then the remaining 13 coins are worth 65 cents. Since each of the remaining coins is worth at least 5 cents, this is possible only if each of the 13 coins is a nickel. So one combination that works is 35 pennies and 13 nickels.

If there are 40 pennies, then the remaining 8 coins are worth 60 cents.

We now look at the number of quarters in this combination.

If there are 0 quarters, then we must have 8 nickels and dimes totalling 60 cents. If all of the 8 coins were nickels, they would be worth 40 cents, so we need to change 4 nickels to dimes to increase our total by 20 cents to 60 cents. Therefore, 40 pennies, 0 quarters, 4 nickels and 4 dimes works.

If there is 1 quarter, then we must have 7 nickels and dimes totalling 35 cents. Since each remaining coin is worth at least 5 cents, then all of the 7 remaining coins must be nickels. Therefore, 40 pennies, 1 quarter, 7 nickels and 0 dimes works.

If there are 2 quarters, then we must have 6 nickels and dimes totalling 10 cents. This is impossible. If there were more than 2 quarters, the quarters would be worth more than 60 cents, so this is not possible.

If there are 45 pennies, then the remaining 3 coins are worth 55 cents in total.

In order for this to be possible, there must be 2 quarters (otherwise the maximum value of the 3 coins would be with 1 quarter and 2 dimes, or 45 cents).

This means that the remaining coin is worth 5 cents, and so is a nickel.

Therefore, 45 pennies, 2 quarters, 1 nickel and 0 dimes is a combination that works.

Therefore, there are 4 combinations that work.