AMC 10 · 2004 · #2
Grade 7 arithmeticFor any three real numbers a, b, and c, with b=c, the operation ⊗ is defined by:
⊗(a,b,c)=b−ca
What is ⊗(⊗(1,2,3),⊗(2,3,1),⊗(3,1,2))?
Pick an answer.
AMC 10 2004 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: A three-input operation is defined by $\otimes(a,b,c)=\dfrac{a}{b-c}$, valid whenever $b\neq c$. Evaluate the nested expression $\otimes\big(\otimes(1,2,3),\,\otimes(2,3,1),\,\otimes(3,1,2)\big)$.
Givens: The rule $\otimes(a,b,c)=\dfrac{a}{b-c}$: the first input goes on top, and the second minus the third goes on the bottom.; The three inner calls are $\otimes(1,2,3)$, $\otimes(2,3,1)$, and $\otimes(3,1,2)$.; Answer choices: (A) $-\tfrac{1}{2}$, (B) $-\tfrac{1}{4}$, (C) $0$, (D) $\tfrac{1}{4}$, (E) $\tfrac{1}{2}$.
Unknowns: The single number that the whole nested expression equals.
Understand
Restated: A three-input operation is defined by $\otimes(a,b,c)=\dfrac{a}{b-c}$, valid whenever $b\neq c$. Evaluate the nested expression $\otimes\big(\otimes(1,2,3),\,\otimes(2,3,1),\,\otimes(3,1,2)\big)$.
Givens: The rule $\otimes(a,b,c)=\dfrac{a}{b-c}$: the first input goes on top, and the second minus the third goes on the bottom.; The three inner calls are $\otimes(1,2,3)$, $\otimes(2,3,1)$, and $\otimes(3,1,2)$.; Answer choices: (A) $-\tfrac{1}{2}$, (B) $-\tfrac{1}{4}$, (C) $0$, (D) $\tfrac{1}{4}$, (E) $\tfrac{1}{2}$.
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
The expression is one operation wrapped around three copies of itself, so it splits into clean subproblems (Tool #7): first evaluate the three inner $\otimes$ calls, then feed those three numbers into the outer $\otimes$. The whole trap is signs — two inner calls come out negative, and the outer bottom is a subtraction of a negative. Tool #3 (Eliminate Possibilities) closes it: once the result is a negative fraction, the positive choices $0$, $\tfrac14$, $\tfrac12$ are gone, and only its size decides between $-\tfrac12$ and $-\tfrac14$.
Execute — Answer: B
6.EE.A.2 Step 1 Decode the operation
- The rule $\otimes(a,b,c)=\dfrac{a}{b-c}$ is a machine: put the first number on top, and the second number minus the third on the bottom.
- Because the expression is $\otimes$ of three other $\otimes$ results, the inner three calls must be worked out first, exactly like parentheses inside parentheses.
💡 A defined operation is just a fill-in-the-blanks formula, so read which slot each number falls into before computing.
7.NS.A.2 Step 2 Evaluate the three inner calls
- Apply the rule to each inner triple.
- $\otimes(1,2,3)=\dfrac{1}{2-3}=\dfrac{1}{-1}=-1$.
- $\otimes(2,3,1)=\dfrac{2}{3-1}=\dfrac{2}{2}=1$.
- $\otimes(3,1,2)=\dfrac{3}{1-2}=\dfrac{3}{-1}=-3$.
- Two of the three come out negative because their bottoms are negative.
💡 A positive divided by a negative is negative, so the sign of each bottom decides the sign of each inner value.
7.NS.A.1 Step 3 Set up the outer call
- The three inner results $-1$, $1$, and $-3$ become the outer inputs in order, so we need $\otimes(-1,1,-3)=\dfrac{-1}{1-(-3)}$.
- The bottom is $1-(-3)$: subtracting $-3$ is the same as adding $3$, so $1-(-3)=1+3=4$.
💡 Subtracting a negative flips it to addition, which is the one place a hidden sign error usually sneaks in.
7.NS.A.2 Step 4 Finish and rule out the traps
- The outer value is $\dfrac{-1}{4}=-\dfrac{1}{4}$: a negative top over a positive bottom stays negative.
- That is choice (B).
- The sign kills the non-negative choices $0$ (C), $\tfrac14$ (D), and $\tfrac12$ (E), and the size rules out $-\tfrac12$ (A), which would need a bottom of $2$, not $4$.
- So the answer is (B).
💡 Nail the sign first and half the choices vanish; then the denominator's size pins down which negative fraction it is.
6.EE.A.2 The rule $\otimes(a,b,c)=\dfrac{a}{b-c}$ is a machine: put the first number on t 7.NS.A.2 Apply the rule to each inner triple. $\otimes(1,2,3)=\dfrac{1}{2-3}=\dfrac{1}{-1 7.NS.A.1 The three inner results $-1$, $1$, and $-3$ become the outer inputs in order, so 7.NS.A.2 The outer value is $\dfrac{-1}{4}=-\dfrac{1}{4}$: a negative top over a positive Review
Reasonableness: The final fraction is $\dfrac{-1}{4}$: a small negative number near $-0.25$. It should be negative because the top ($-1$) is negative and the bottom ($4$) is positive, which immediately rejects the non-negative choices. Its size is a quarter, not a half, so $-\tfrac12$ is too big and $-\tfrac14$ fits. Both checks land on (B).
Alternative: Track only signs first. The inner values are $-1$, $1$, $-3$, so the outer call is $\dfrac{\text{negative}}{\text{positive}}$, which must be negative — that alone eliminates (C), (D), and (E). Then compute just the magnitude, $\dfrac{1}{4}$, to choose (B) over (A). Reaching the same letter by sign-then-size confirms the answer without recomputing the whole expression.
CCSS standards used (min grade 7)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Reading the definition $\otimes(a,b,c)=\frac{a}{b-c}$ as a formula and substituting each triple into the correct slots.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Computing the outer bottom $1-(-3)=4$ by turning subtraction of a negative into addition.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Dividing signed numbers to get the inner values $-1,1,-3$ and the final result $-\frac{1}{4}$.)
⭐ When an operation is nested inside itself, finish every inner piece first, then drop those numbers into the outer rule — and watch the minus signs at each step.
⭐ When an operation is nested inside itself, finish every inner piece first, then drop those numbers into the outer rule — and watch the minus signs at each step.
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