AMC 10 · 2010 · #5
Grade 7 algebraPick an answer.
Keep e as an unknown letter and turn both computations into expressions in e (Tool #4, Introduce a Variable). Tool #7 (Identify Subproblems) handles the correct expression one layer at a time, peeling the nested parentheses from the inside out so the minus signs are distributed without error. Then Tool #13 (Convert to Algebra) turns the sentence "Larry's answer equals the correct answer" into a single equation in e, which solves in one line. Carrying e symbolically is cleaner than testing all five choices, because it produces the value directly.
Expand the correct expression
Expanding the brackets flips some terms twice.
A minus in front of a group flips the sign of everything inside it, so peel one shell at a time to keep the flips straight.
A minus in front of a group flips the sign of everything inside it, so the shells are peeled one at a time.
▸ Why?
A factor placed across a sum reaches every term inside it, and a minus sign is such a factor.
▸ Why?
A term and its opposite add to nothing, so a flipped sign is exactly what makes pairs cancel or survive.
Write Larry's stripped version
The bracket-free version keeps every sign as written.
Ignoring parentheses just means taking every sign at face value, with no flipping.
7.EE.A.1Convert To AlgebraSubstitute the known numbers
Substituting leaves one unknown on each side.
Combining the signed numbers first collapses each side to a simple form ± e plus a constant.
7.NS.A.1Introduce A VariableSet them equal and solve
Setting them equal gives 3, choice (D).
Setting the two expressions equal turns the "lucky coincidence" into one equation with a single solution.
6.EE.B.7Introduce A VariableA minus sign in front of a group flips every sign inside it, so expand the real expression carefully, write Larry's no-parentheses version, set the two equal, and solve for e to get 3.
- Expand the correct expression
- Write Larry's stripped version
- Substitute the known numbers
- Set them equal and solve