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Problem of the Week
Problem D and Solution
Stacks and Stacks


Virat has a large collection of $2 bills and $5 bills. He makes stacks that have a value of $100. Each stack has a least one $2 bill, at least one $5 bill, and no other types of bills. If each stack has a different number of $2 bills than any other stack, what is the maximum number of stacks that Virat can create?


Consider a stack of bills with a total value of $100 that includes \(x\) $2 bills and \(y\) $5 bills. The $2 bills are worth $2\(x\) and the $5 bills are worth $5\(y\), and so \(2x + 5y = 100\).

Determining the number of possible stacks that the teller could have is equivalent to determining the numbers of pairs \((x, y)\) of integers with \(x \geq 1\) and \(y \geq 1\) and \(2x + 5y = 100\) or \(5y=100-2x\). (We must have \(x \geq 1\) and \(y \geq 1\) because each stack includes at least one $2 bill and at least one $5 bill.)

Since \(x \geq 1\), then:

\[\begin{aligned} 2x &\geq 2\\ 2x + 98 &\geq 100\\ 98 &\geq 100 - 2x \\\end{aligned}\]

This can be rewritten as \(100 - 2x\leq 98\).

Also, since \(5y = 100-2x\), this becomes \(5y \leq 98\).

This means that \(y \leq \frac{98}{5} = 19.6\). Since \(y\) is an integer, then \(y \leq 19\).

Notice that since \(5y = 100 - 2x\), then the right side is the difference between two even integers and is therefore even. This means that \(5y\) (the left side) is itself even, which means that \(y\) must be even.

Since \(y\) is even, \(y \geq 1\), and \(y \leq 19\), then the possible values of \(y\) are 2, 4, 6, 8, 10, 12, 14, 16, 18.

Each of these values gives a pair \((x, y)\) that satisfies the equation \(2x + 5y = 100\). These ordered pairs are:

\((x, y) = (45, 2),(40, 4),(35, 6),(30, 8),(25, 10),(20, 12),(15, 14),(10, 16),(5, 18)\).

Therefore, we see that the maximum number of stacks that Virat could have is 9.