# 2005 CEMC Gauss (Grade 7) Problems/Problem 25

## Problem

How many different combinations of pennies, nickels, dimes and quarters use $48$ coins to total $1$ dollar?

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

## 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. Thus, the answer is $B$.

## 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 totaling $60$ 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 totaling $35$ 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 totaling $10$ 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. The answer is $B$.