AMC 8 · 2000 · #17

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition

fraction-arithmeticfunction-evaluationorder-of-operations identify-subproblemsformula-substitution ↑ Prerequisites: fraction-arithmeticorder-of-operations

📏 Medium solution 💡 3 insights

Problem

The operation $\otimes$ is defined for all nonzero numbers by $a\otimes b =\frac{a^{2}}{b}$ . Determine $[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)]$ .

Pick an answer.

(A)

$-rac{2}{3}$

(B)

$-rac{1}{4}$

(C)

(D)

$\frac{1}{4}$

(E)

$\frac{2}{3}$

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Toolkit + CCSS Solution

Understand

Restated: A new operation is defined by $a \otimes b = \dfrac{a^{2}}{b}$ for all nonzero numbers. Compute $[(1 \otimes 2) \otimes 3] - [1 \otimes (2 \otimes 3)]$ — the same three numbers, but the brackets group them differently on the two sides.

Givens: Rule: $a \otimes b = \dfrac{a^{2}}{b}$; Expression to evaluate: $[(1 \otimes 2) \otimes 3] - [1 \otimes (2 \otimes 3)]$; Answer choices: (A) $-\dfrac{2}{3}$, (B) $-\dfrac{1}{4}$, (C) $0$, (D) $\dfrac{1}{4}$, (E) $\dfrac{2}{3}$

Unknowns: The value of $[(1 \otimes 2) \otimes 3] - [1 \otimes (2 \otimes 3)]$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #3 Eliminate Possibilities

The bracket structure splits the work into four small substitutions — Tool #7 (Identify Subproblems) handles it cleanly. Evaluate the inner $\otimes$ on each side, then feed the result into the outer $\otimes$, then subtract. Because $\otimes$ is non-associative (the left input gets squared, the right one does not), the two sides will land on different numbers; the difference is the answer. Tool #3 (Eliminate Possibilities) is the AMC multiple-choice safety net: a quick sign check (left side is small, right side is close to $1$, so the difference must be negative) already rules out (C), (D), (E).

Execute — Answer: A

#7 Identify Subproblems 5.OA.A.1 Step 1

Split the problem into four substitutions.
Left side: first $1 \otimes 2$, then that result $\otimes 3$.
Right side: first $2 \otimes 3$, then $1 \otimes$ that result.
Finally, subtract.

$$[(1 \otimes 2) \otimes 3] - [1 \otimes (2 \otimes 3)]$$

💡 Reading the brackets first is the Grade 5 "use parentheses in numerical expressions" rule, and it organises the work into four bite-sized steps.

#7 Identify Subproblems 6.EE.A.2 Step 2

Left side, inner $\otimes$.
Apply the rule with $a = 1$, $b = 2$: square the $1$, divide by $2$.

$$1 \otimes 2 \;=\; \dfrac{1^{2}}{2} \;=\; \dfrac{1}{2}$$

💡 Plugging numbers into the letters $a$ and $b$ of a defined formula is the Grade 6 "evaluate expressions where letters stand for numbers" move.

#7 Identify Subproblems 6.NS.A.1 Step 3

Left side, outer $\otimes$.
Feed $\tfrac{1}{2}$ into the rule as $a$, with $b = 3$: square $\tfrac{1}{2}$ to get $\tfrac{1}{4}$, then divide by $3$.

$$\left(\dfrac{1}{2}\right) \otimes 3 \;=\; \dfrac{(1/2)^{2}}{3} \;=\; \dfrac{1/4}{3} \;=\; \dfrac{1}{12}$$

💡 Dividing the fraction $\tfrac{1}{4}$ by the whole number $3$ multiplies the denominator: $\tfrac{1}{4 \cdot 3} = \tfrac{1}{12}$.

#7 Identify Subproblems 6.EE.A.2 Step 4

Right side, inner $\otimes$.
Apply the rule with $a = 2$, $b = 3$: square the $2$ to get $4$, then divide by $3$.

$$2 \otimes 3 \;=\; \dfrac{2^{2}}{3} \;=\; \dfrac{4}{3}$$

💡 Same substitution move as before — this time the left input is $2$, so squaring matters and the result jumps to $4$.

#7 Identify Subproblems 6.NS.A.1 Step 5

Right side, outer $\otimes$.
Feed $\tfrac{4}{3}$ in as $b$, with $a = 1$: square the $1$, then divide by $\tfrac{4}{3}$.
Dividing by $\tfrac{4}{3}$ is the same as multiplying by its reciprocal $\tfrac{3}{4}$.

$$1 \otimes \dfrac{4}{3} \;=\; \dfrac{1^{2}}{4/3} \;=\; 1 \cdot \dfrac{3}{4} \;=\; \dfrac{3}{4}$$

💡 "Divide by a fraction" $=$ "multiply by its reciprocal" — the Grade 6 fraction-division habit.

#7 Identify Subproblems 5.NF.A.1 Step 6

Subtract the right-side value from the left-side value.
Use a common denominator of $12$.

$$\dfrac{1}{12} - \dfrac{3}{4} \;=\; \dfrac{1}{12} - \dfrac{9}{12} \;=\; -\dfrac{8}{12} \;=\; -\dfrac{2}{3}$$

💡 Rewriting $\tfrac{3}{4}$ as $\tfrac{9}{12}$ is the Grade 5 "common denominator" step, then simplify $-\tfrac{8}{12}$ by dividing top and bottom by $4$.

#3 Eliminate Possibilities 5.OA.A.1 Step 7

Match the result to the answer choices.
Only choice (A) $-\tfrac{2}{3}$ fits.
Sign-check the other choices: the right side $\tfrac{3}{4}$ is much bigger than the left side $\tfrac{1}{12}$, so the difference is negative — that already eliminates (C), (D), (E).
And (B) $-\tfrac{1}{4}$ misses the magnitude.

$$-\dfrac{2}{3} \;\Rightarrow\; \textbf{(A)}$$

💡 Lining the computed value up against the five choices is the AMC multiple-choice habit and catches sign or magnitude slips.

[1] #7 5.OA.A.1 Split the problem into four substitutions. Left side: first $1 \otimes 2$, then

[2] #7 6.EE.A.2 Left side, inner $\otimes$. Apply the rule with $a = 1$, $b = 2$: square the $1$

[3] #7 6.NS.A.1 Left side, outer $\otimes$. Feed $\tfrac{1}{2}$ into the rule as $a$, with $b =

[4] #7 6.EE.A.2 Right side, inner $\otimes$. Apply the rule with $a = 2$, $b = 3$: square the $2

[5] #7 6.NS.A.1 Right side, outer $\otimes$. Feed $\tfrac{4}{3}$ in as $b$, with $a = 1$: square

[6] #7 5.NF.A.1 Subtract the right-side value from the left-side value. Use a common denominator

[7] #3 5.OA.A.1 Match the result to the answer choices. Only choice (A) $-\tfrac{2}{3}$ fits. Si

Review

Reasonableness: Sanity check the sign and size. The right-side answer $\tfrac{3}{4}$ is close to $1$ because dividing $1$ by something a little bigger than $1$ (namely $\tfrac{4}{3}$) gives a number a little smaller than $1$. The left-side answer $\tfrac{1}{12}$ is much smaller because we squared a fraction ($\tfrac{1}{2} \to \tfrac{1}{4}$) and then divided by another whole number. So left $-$ right is small minus near-$1$ — clearly negative and not tiny, which matches $-\tfrac{2}{3}$. The non-zero answer also confirms that $\otimes$ is not associative; if it were, the difference would be $0$ (choice (C)), so the problem is really testing that observation.

Alternative: Tool #9 (Solve an Easier Related Problem): rewrite $\otimes$ as an ordinary function $f(a, b) = \dfrac{a^{2}}{b}$. The expression becomes $f(f(1, 2),\, 3) - f(1,\, f(2, 3))$, which is the standard "is this function associative?" question. Compute $f(1, 2) = \tfrac{1}{2}$ and $f(\tfrac{1}{2}, 3) = \tfrac{1}{12}$; then $f(2, 3) = \tfrac{4}{3}$ and $f(1, \tfrac{4}{3}) = \tfrac{3}{4}$. Subtract: $\tfrac{1}{12} - \tfrac{3}{4} = -\tfrac{2}{3}$, again (A).

CCSS standards used (min grade 6)

5.OA.A.1 Use parentheses, brackets, or braces in numerical expressions and evaluate (Respecting the brackets to evaluate each inner $\otimes$ before the outer $\otimes$, and matching the final value against the multiple-choice list.)
6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Substituting numerical values into the letters $a$ and $b$ of the rule $\dfrac{a^{2}}{b}$ on each of the four applications of $\otimes$.)
6.NS.A.1 Apply and extend previous understandings of multiplication and division to divide fractions (Computing $\tfrac{1/4}{3} = \tfrac{1}{12}$ on the left side and $\dfrac{1}{4/3} = \tfrac{3}{4}$ on the right side — both are fraction divisions.)
5.NF.A.1 Add and subtract fractions with unlike denominators by finding a common denominator (Rewriting $\tfrac{3}{4} = \tfrac{9}{12}$ to subtract $\tfrac{1}{12} - \tfrac{9}{12} = -\tfrac{8}{12} = -\tfrac{2}{3}$.)

⭐ Brackets pick which number gets squared, and squaring a fraction shrinks it while squaring a whole number grows it — that asymmetry is exactly why the two sides do not match, and the difference comes out to $-\tfrac{2}{3}$, answer (A).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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