2003 AMC 10A Problem 13

The second and third numbers appear together ("the sum of the other two"), so treat that sum as one block first.

1. Call the three numbers $a$, $b$, $c$. Let the block be $s = b + c$.
2. The first number is four times the block: $a = 4s$.
3. All three add to $20$, so $a + s = 20$. Replace $a$ by $4s$:
$$4s + s = 20, \qquad 5s = 20, \qquad s = 4.$$
4. So $a = 4 \times 4 = 16$, and $b + c = 4$.
5. Now split the $4$. The second is seven times the third, so $b = 7c$ and
$$7c + c = 4, \qquad 8c = 4, \qquad c = \frac{1}{2}.$$
Then $b = 7 \times \frac{1}{2} = \frac{7}{2}$.
6. Check: $16 + \frac{7}{2} + \frac{1}{2} = 16 + 4 = 20$. Correct.
7. The product is
$$16 \times \frac{7}{2} \times \frac{1}{2} = 16 \times \frac{7}{4} = 28.$$

The answer is $\boxed{\textbf{(A)}\ 28}$.

"One number is $4$ times the rest" means the total is split into $4 + 1 = 5$ equal parts, and that number gets $4$ of them. Treating $b + c$ as one block turns three unknowns into one easy split, and then a second split finishes the job. Look for phrases like "the sum of the others" and treat that sum as a single unknown.

Call the numbers $a, b, c$. We have $a + b + c = 20$, $a = 4(b + c)$ and $b = 7c$.

Let $s = b + c$. Then $a = 4s$ and $a + s = 20$ give $5s = 20$, so $s = 4$ and $a = 16$.

Now $b + c = 4$ with $b = 7c$ gives $8c = 4$, so $c = \tfrac{1}{2}$ and $b = \tfrac{7}{2}$.

The product is $16 \cdot \tfrac{7}{2} \cdot \tfrac{1}{2} = 28$.

The answer is $\boxed{\textbf{(A)}\ 28}$.