Virat has a large collection of \(\$2\) bills and \(\$5\) bills. He makes stacks of bills that each have a total value of exactly \(\$100\). 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 \(\$2x\) and the \(\$5\) bills are worth \(\$5y\), 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 \[\begin{aligned} 2x &\geq 2\\ -2x&\leq -2\\ 100 - 2x &\leq 100-2 \\ 100 - 2x &\leq 98 \end{aligned}\]
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 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 \(2\), \(4\), \(6\), \(8\), \(10\), \(12\), \(14\), \(16\), and \(18.\)
Each of these values gives a pair \((x, y)\) that satisfies the equation \(2x + 5y = 100\). These ordered pairs are \((45, 2)\), \((40, 4)\), \((35, 6)\), \((30, 8)\), \((25, 10)\), \((20, 12)\), \((15, 14)\), \((10, 16)\), and \((5, 18).\)
Therefore, the maximum number of stacks that Virat can create is \(9.\)