AMC 10 · 2010 · #9
Grade 7 algebraLucky 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 parenthese but added and subtracted correctly and obtained the correct result by coincidence. The number Larry substituted for a, b, c, and d were 1, 2, 3, and 4, respectively. What number did Larry substitute for e?
Pick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The expression is $a-(b-(c-(d+e)))$. Larry drops every set of parentheses but keeps each $+$ and $-$ sign exactly where it is written, then adds and subtracts left to right. By luck his stripped-down answer matches the real value of the full expression. With $a=1$, $b=2$, $c=3$, $d=4$, we need the value of $e$ that makes the two results agree.
Givens: The correct expression is $a-(b-(c-(d+e)))$.; Larry writes the same numbers and signs but ignores all parentheses: $a-b-c-d+e$.; Larry's answer equals the correct answer.; $a=1,\ b=2,\ c=3,\ d=4$.; Answer choices: (A) $-5$, (B) $-3$, (C) $0$, (D) $3$, (E) $5$.
Unknowns: The value of $e$ that Larry substituted.
Understand
Restated: The expression is $a-(b-(c-(d+e)))$. Larry drops every set of parentheses but keeps each $+$ and $-$ sign exactly where it is written, then adds and subtracts left to right. By luck his stripped-down answer matches the real value of the full expression. With $a=1$, $b=2$, $c=3$, $d=4$, we need the value of $e$ that makes the two results agree.
Givens: The correct expression is $a-(b-(c-(d+e)))$.; Larry writes the same numbers and signs but ignores all parentheses: $a-b-c-d+e$.; Larry's answer equals the correct answer.; $a=1,\ b=2,\ c=3,\ d=4$.; Answer choices: (A) $-5$, (B) $-3$, (C) $0$, (D) $3$, (E) $5$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #7 Identify Subproblems
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.
Execute — Answer: D
5.OA.A.1 Step 1 Expand the correct expression
- Remove the parentheses in $a-(b-(c-(d+e)))$ from the inside out, flipping every sign that sits behind a minus.
- Innermost: $c-(d+e)=c-d-e$.
- Next: $b-(c-d-e)=b-c+d+e$.
- Outermost: $a-(b-c+d+e)=a-b+c-d-e$.
💡 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.
7.EE.A.1 Step 2 Write Larry's stripped version
- Larry keeps the numbers and their written signs but throws away the parentheses, reading straight across: $a-b-c-d+e$.
- Only the final $+e$ and the leading $a$ stay positive; the middle terms are all subtracted.
💡 Ignoring parentheses just means taking every sign at face value, with no flipping.
7.NS.A.1 Step 3 Substitute the known numbers
- Put $a=1$, $b=2$, $c=3$, $d=4$ into both expressions, keeping $e$ as a letter.
- Correct value: $1-2+3-4-e=-2-e$.
- Larry's value: $1-2-3-4+e=-8+e$.
💡 Combining the signed numbers first collapses each side to a simple form $\pm e$ plus a constant.
6.EE.B.7 Step 4 Set them equal and solve
- The coincidence says the two values match: $-2-e=-8+e$.
- Add $e$ to both sides and add $8$ to both sides to get $6=2e$, so $e=3$.
- Both sides then equal $-5$, confirming the match.
- The answer is $\textbf{(D)}\ 3$.
💡 Setting the two expressions equal turns the "lucky coincidence" into one equation with a single solution.
5.OA.A.1 Remove the parentheses in $a-(b-(c-(d+e)))$ from the inside out, flipping every 7.EE.A.1 Larry keeps the numbers and their written signs but throws away the parentheses, 7.NS.A.1 Put $a=1$, $b=2$, $c=3$, $d=4$ into both expressions, keeping $e$ as a letter. C 6.EE.B.7 The coincidence says the two values match: $-2-e=-8+e$. Add $e$ to both sides an Review
Reasonableness: Plug $e=3$ back into the original problem. Correct: $1-(2-(3-(4+3)))=1-(2-(3-7))=1-(2-(-4))=1-6=-5$. Larry: $1-2-3-4+3=-5$. Both give $-5$, so the coincidence really happens and $e=3$, choice (D), is right.
Alternative: Skip substituting until the end. The two expanded expressions are $a-b+c-d-e$ and $a-b-c-d+e$. Their difference is $(a-b+c-d-e)-(a-b-c-d+e)=2c-2e$. For the results to be equal this difference must be $0$, so $2c-2e=0$, giving $e=c=3$ regardless of $a$, $b$, and $d$. This shows Larry's trick works exactly when $e$ equals $c$.
CCSS standards used (min grade 7)
5.OA.A.1Use parentheses, brackets, or braces in numerical expressions and evaluate (Peeling the nested parentheses in the correct expression from the inside out and respecting the grouping they force.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Rewriting the expression without parentheses to capture what Larry actually computed.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Combining the signed numbers $1,-2,+3,-4$ (and $\pm e$) to reduce each side to a constant plus $e$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $6=2e$ to find $e=3$.)
⭐ A 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 $3$.
⭐ A 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 $3$.
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