2010 AMC 10B Problem 9

P1-10~2 minPrint

Algebraic Manipulation Linear Equations & Word Problems Set up the equation/formula and compute; no special trick needed

Lucky Larry's teacher asked him to substitute numbers for $a$ , $b$ , $c$ , $d$ , and $e$ in the expression $a-(b-(c-(d+e)))$ and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for $a$ , $b$ , $c$ , and $d$ were $1$ , $2$ , $3$ , and $4$ , respectively. What number did Larry substitute for $e$ ?

Show solution without answering

Key insight

Plug in the known numbers first. The correct value works out to -2 - e and Larry's value is e - 8; setting them equal gives e = 3.

Solution

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 $\boxed{\textbf{(D)}\ 3}$ .

Why this works

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.

Alternative approach

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 $e = c = 3$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(D)}\ 3}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 22: Special Manipulations
Introduction to Algebra, Chapter 7: Proportion (§7.4)