Problem of the Week
Problem B and Solution
Ask the Banker

Problem

Vesna is the banker in a board game that uses $\mathrm{\$}1$, $\mathrm{\$}10$, and $\mathrm{\$}100$ bills.

1. Vesna needs to give a player $\mathrm{\$}2163$. How can you do this using the fewest total number of bills? How can you do this using the greatest total number of bills?

2. Is it possible to give a player $\mathrm{\$}254$ using exactly $20$ bills in total? How about using exactly $30$ bills in total? If so, show how it’s possible. If not, explain why it’s not possible.

Extension: Vesna likes when she can give a player the same number of each type of bill. For which total amounts of money is this possible? Explain.

Solution

1. To give a player $\mathrm{\$}2163$ using the fewest total number of bills, we first use as many $\mathrm{\$}100$ bills as we can. Since the number $2163$ has $21$ hundreds, then we can use at most $21$ of the $\mathrm{\$}100$ bills. This gives $21\times \mathrm{\$}100=\mathrm{\$}2100$, so we are left with $\mathrm{\$}2163-\mathrm{\$}2100=\mathrm{\$}63$. Next we use as many $\mathrm{\$}10$ bills as we can. Since the number $63$ has $6$ tens, then we can use at most $6$ of the $\mathrm{\$}10$ bills. This gives $6\times \mathrm{\$}10=\mathrm{\$}60$. We are then left with $\mathrm{\$}3$, so we need $3$ of the $\mathrm{\$}1$ bills. Thus in total, we use: $$21\times \mathrm{\$}100\text{ bills};6\times \mathrm{\$}10\text{ bills};3\times \mathrm{\$}1\text{ bills}$$ This is a total of $21+6+3=30$ bills.

   To give a player $\mathrm{\$}2163$ using the greatest total number of bills, we want to use as many $\mathrm{\$}1$ bills as possible. If we use all $\mathrm{\$}1$ bills, then we will use $2163$ bills in total.

2. Using the strategy from (a) to use the fewest total number of bills, we can give $\mathrm{\$}254$ as follows: $$2\times \mathrm{\$}100\text{ bills};5\times \mathrm{\$}10\text{ bills};4\times \mathrm{\$}1\text{ bills}$$ This uses a total of $2+5+4=11$ bills. Let’s try replacing one $\mathrm{\$}100$ bill with ten $\mathrm{\$}10$ bills. This gives: $$1\times \mathrm{\$}100\text{ bill};15\times \mathrm{\$}10\text{ bills};4\times \mathrm{\$}1\text{ bills}$$ This uses a total of $1+15+4=20$ bills, so it is possible to give a player $\mathrm{\$}254$ using exactly $20$ bills in total.

   Notice that every time we replace a bill with ten smaller bills, the total number of bills increases by $9$. This is true if we replace one $\mathrm{\$}100$ bill with ten $\mathrm{\$}10$ bills or if we replace one $\mathrm{\$}10$ bill with ten $\mathrm{\$}1$ bills. So the total number of bills is a sequence that starts at $11$ and increases by $9$ each time, until it reaches $254$ (which is the greatest total number of bills that can be used for $\mathrm{\$}254$). Writing out more terms in the sequence gives $11$, $20$, $29$, $38$, $\dots$. Since $30$ is not in this sequence, we can not give a player $\mathrm{\$}254$ using exactly $30$ bills in total.

Solution to Extension: If a player gets $1$ of each bill, then the total amount is $1\times \mathrm{\$}100+1\times \mathrm{\$}10+1\times \mathrm{\$}1=\mathrm{\$}111$. If a player gets $2$ of each bill, then the total amount will be $2\times \mathrm{\$}111=\mathrm{\$}222$, because they have twice as many of each bill, so the total amount will double. Similarly, if a player gets $6$ of each bill, then the total amount will be $6\times \mathrm{\$}111=\mathrm{\$}666$, and if a player gets $15$ of each bill, then the total amount will be $15\times \mathrm{\$}111=\mathrm{\$}1665$. Thus, the total amount will always be a multiple of $111$.

We can also use variables to explain this. Suppose a player receives $n$ bills of each type. Then the total amount is equal to $(n\times 100+n\times 10+n\times 1)$. This is the same as $n\times (100+10+1)$, which equals $n\times 111$. Therefore, the total amount must be a multiple of $111$.