2010 AMC 10B Problem 9

We know $a, b, c, d$, so put them in right away. Then the only unknown is $e$.

1. The correct value. Work from the innermost parentheses outward.
- Innermost: $3 - (4 + e) = 3 - 4 - e = -1 - e$.
- Next: $2 - (-1 - e) = 2 + 1 + e = 3 + e$.
- Outer: $1 - (3 + e) = 1 - 3 - e = -2 - e$.
2. Larry's value. He dropped every parenthesis and read the signs left to right:
$$1 - 2 - 3 - 4 + e = e - 8.$$
3. They are equal, so
$$-2 - e = e - 8.$$
Add $e$ and $8$ to both sides: $6 = 2e$, so $e = 3$.
4. Check: the correct value is $-2 - 3 = -5$ and Larry's is $3 - 8 = -5$. They match.

The answer is $\boxed{\textbf{(D)}\ 3}$.

Nested parentheses are easiest from the inside out, one layer at a time. The key rule is that a minus sign in front of parentheses flips the sign of every term inside. Once both values are written in terms of $e$ alone, the problem is a one-step equation.

Compare the two expressions before plugging in numbers. Removing the parentheses gives
$$a - (b - (c - (d + e))) = a - b + c - d - e,$$
while Larry computed $a - b - c - d + e$. The terms $a$, $-b$ and $-d$ are the same in both. The only difference is $+c - e$ against $-c + e$, and these are equal exactly when $e = c = 3$.

Remove the parentheses from the inside out, flipping signs each time a minus sign is distributed:
$$c - (d+e) = c - d - e,\qquad b - (c - d - e) = b - c + d + e,$$
$$a - (b - c + d + e) = a - b + c - d - e.$$

Larry, ignoring parentheses, computed $a - b - c - d + e$.

The two results are equal when
$$a - b + c - d - e = a - b - c - d + e \quad\Longrightarrow\quad 2c = 2e \quad\Longrightarrow\quad e = c.$$
Since $c = 3$, Larry used $e = 3$. (Check: the correct value is $1 - 2 + 3 - 4 - 3 = -5$, and Larry's is $1 - 2 - 3 - 4 + 3 = -5$.)

The answer is $\boxed{\textbf{(D)}\ 3}$.