2003 AMC 10B Problem 2

1. Two weeks is $14$ days, and every day Al buys the same thing: one green and one pink pill. So first find one day's cost:
$$546 \div 14 = 39 \text{ dollars}.$$
2. So a green pill and a pink pill together cost $39$ dollars, and the green one costs $1$ dollar more.
3. If the two prices were equal, each would be $19.50$. The green is $50$ cents above that and the pink is $50$ cents below, so green $= 20$ and pink $= 19$. (With algebra: pink $= p$, green $= p+1$, so $2p + 1 = 39$ and $p = 19$.)
4. Check: $20 + 19 = 39$, and $14 \times 39 = 546$.

The answer is $\boxed{\textbf{(D)}\ \$20}$.

The total is the same daily purchase repeated $14$ times, so dividing by $14$ shrinks the problem to one day. Then it is a "sum and difference" puzzle: the bigger number is (sum plus difference) divided by $2$, here $(39+1)/2 = 20$. When a total comes from repeating the same purchase, find the cost of one repeat first.

Two weeks is $14$ days, and each day Al buys exactly one pill of each color. The daily cost of the pair is therefore
$$\frac{546}{14} = 39 \text{ dollars}.$$

Let the pink pill cost $p$ dollars; the green pill costs $p + 1$. Then $p + (p+1) = 39$, so $2p = 38$ and $p = 19$. The green pill costs $19 + 1 = 20$ dollars.

Check: $14 \cdot (20 + 19) = 14 \cdot 39 = 546$.

The answer is $\boxed{\textbf{(D)}\ \$20}$.