AMC 8 · 2000 · #17

Grade 6 arithmetic
Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation
fraction-arithmeticfunction-evaluationorder-of-operations identify-subproblemsformula-substitution ↑ Prerequisites: fraction-arithmeticorder-of-operations
📏 Medium solution 💡 3 insights
Problem
A new operation is defined by a ⊗ b = a2b\frac{a²}{b} for all nonzero numbers. Compute [(1 ⊗ 2) ⊗ 3] - [1 ⊗ (2 ⊗ 3)] — the same three numbers, but the brackets group them differently on the two sides.

Pick an answer.

(A)
$-\frac{2}{3}$
(B)
$-\frac{1}{4}$
(C)
0
(D)
$\frac{1}{4}$
(E)
$\frac{2}{3}$

AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The bracket structure splits the work into four small substitutions — Tool #7 (Identify Subproblems) handles it cleanly. Evaluate the inner ⊗ on each side, then feed the result into the outer ⊗, then subtract. Because ⊗ 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).

1STEP 1

Split into four substitutions: each side's inner ⊗ first, then its outer ⊗, then subtract left minus right.

[(1 ⊗ 2) ⊗ 3] - [1 ⊗ (2 ⊗ 3)]
2STEP 2

Left inner ⊗: with a = 1, b = 2, square 1 and divide by 2 to get 12\frac{1}{2}.

1 ⊗ 2 = 122\frac{1²}{2} = 12\frac{1}{2}
3STEP 3

Left outer ⊗: feed 12\frac{1}{2} in as a with b = 3, square to 14\frac{1}{4}, then divide by 3 to get 112\frac{1}{12}.

(12\frac{1}{2}) ⊗ 3 = (12)23\frac{(\frac{1}{2})²}{3} = 143\frac{\frac{1}{4}}{3} = 112\frac{1}{12}
4STEP 4

Right inner ⊗: with a = 2, b = 3, square 2 to get 4, then divide by 3, giving 43\frac{4}{3}.

2 ⊗ 3 = 223\frac{2²}{3} = 43\frac{4}{3}
5STEP 5

Right outer ⊗: square 1, then divide by 43\frac{4}{3} — the same as multiplying by its reciprocal 34\frac{3}{4} — giving 34\frac{3}{4}.

1 ⊗ 43\frac{4}{3} = 1243\frac{1²}{\frac{4}{3}} = 1 · 34\frac{3}{4} = 34\frac{3}{4}
6STEP 6

Subtract left minus right over common denominator 12: 112\frac{1}{12} - 912\frac{9}{12} = −23-\frac{2}{3}.

112\frac{1}{12} - 34\frac{3}{4} = 112\frac{1}{12} - 912\frac{9}{12} = −812-\frac{8}{12} = −23-\frac{2}{3}
7STEP 7

Match −23-\frac{2}{3} to the choices: only (A) fits; a quick sign check (right 34\frac{3}{4} beats left 112\frac{1}{12}) already kills (C), (D), (E).

−23-\frac{2}{3} → (A)
Answer
−23-\frac{2}{3}
Sanity check the sign and size. The right-side answer 34\frac{3}{4} is close to 1 because dividing 1 by something a little bigger than 1 (namely 43\frac{4}{3}) gives a number a little smaller than 1. The left-side answer 112\frac{1}{12} is much smaller because we squared a fraction (12\frac{1}{2} → 14\frac{1}{4}) and then divided by another whole number. So left - right is small minus near-1 — clearly negative and not tiny, which matches −23-\frac{2}{3}. The non-zero answer also confirms that ⊗ is not associative; if it were, the difference would be 0 (choice (C)), so the problem is really testing that observation.
💡Key takeaway

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 −23-\frac{2}{3}, answer (A).

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