2002 AMC 10A Problem 16

Let $x$ be the common value of all five expressions. Then
$$a = x - 1,\quad b = x - 2,\quad c = x - 3,\quad d = x - 4,$$
so $a + b + c + d = 4x - 10$.

The last expression says $a + b + c + d + 5 = x$, i.e. $a + b + c + d = x - 5$. Equating the two descriptions of the sum:
$$4x - 10 = x - 5 \quad\Longrightarrow\quad 3x = 5 \quad\Longrightarrow\quad x = \frac{5}{3}.$$

Therefore $a + b + c + d = x - 5 = \frac{5}{3} - 5 = -\frac{10}{3}$.

The answer is $\boxed{\textbf{(B)}\ -10/3}$.

A chain of equalities is best handled by naming the shared value; every unknown then becomes a simple shift of that one variable. The self-referential last term (the sum itself plus $5$) supplies the single equation needed. Finish by answering the question actually asked, not the intermediate variable.

Compare each expression with the last one: $a + 1 = a + b + c + d + 5$ gives $b + c + d = -4$; similarly $a + c + d = -3$, $a + b + d = -2$, $a + b + c = -1$. Adding all four, $3(a + b + c + d) = -10$, so the sum is $-\frac{10}{3}$.

• Solving for the common value x = 5/3 (choice D) and forgetting that the question asks for a + b + c + d = x - 5.
• Sign slip when writing $a = x - 1$ as $a = x + 1$, which produces $4x + 10 = x - 5$ and the wrong sum $-10$ or $-5$.