AMC 10 · 2011 · #15

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

function-evaluationformula-substitutionlogical-deduction casework ↑ Prerequisites: function-evaluation

📏 Medium solution 💡 3 insights

Problem

A new operation @ means "average with":

a @ b = \frac{a+b}{2}

for all numbers a and b. Three distributive-style laws are proposed: I.

x @ (y + z) = (x @ y) + (x @ z)

; II.

x + (y @ z) = (x + y) @ (x + z)

; III.

x @ (y @ z) = (x @ y) @ (x @ z)

. Decide which of the three hold for every choice of numbers x, y, and z.

Pick an answer.

(A)

I only

(B)

II only

(C)

III only

(D)

I and III only

(E)

II and III only

AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

The symbol @ is just shorthand, so Tool #13 (Convert to Algebra) replaces every @ with its definition (a+b)/2 and turns each law into a plain algebra statement. Tool #7 (Identify Subproblems) splits the job into three independent checks — one per law. For each, simplify both sides and see if they are the same expression; Tool #3 (Eliminate Possibilities) then crosses off any law whose two sides differ, leaving only the true ones.

1STEP 1

Rewrite the operation

Swap every @ for its definition $a @ b = \frac{a+b}{2}$ , then simplify each side of a law into one fraction and compare.

a @ b = (a+b)/2

2STEP 2

Test Law I

Law I: the left side is $\frac{x+y+z}{2}$ but the right side is $\frac{2x+y+z}{2}$ — one extra x, so it fails.

(x+y+z)/2 ≠ (2x+y+z)/2

3STEP 3

Test Law II

Law II: both $x + \frac{y+z}{2}$ and $\frac{(x+y)+(x+z)}{2}$ collapse to $\frac{2x+y+z}{2}$ , so it holds.

(2x+y+z)/2 = (2x+y+z)/2

4STEP 4

Test Law III

Law III: nesting the averages either way flattens to $\frac{2x+y+z}{4}$ on both sides, so it holds too.

(2x+y+z)/4 = (2x+y+z)/4

5STEP 5

Collect the survivors

Cross off Law I; the two laws whose sides matched are the ones that survive, which is choice (E).

I: fails, II: holds, III: holds → (E)

Answer

II and III only

Plug in easy numbers to confirm. Take x=2, y=4, z=6. Law I: 2 @ 10 = 6 but (2@4)+(2@6)=3+4=7 — unequal, so I really fails. Law II: 2 + (4@6) = 2+5 = 7 and (2+4)@(2+6)=6@8=7 — equal. Law III: 2 @ (4@6) = 2@5 = 3.5 and (2@4)@(2@6)=3@4=3.5 — equal. Numbers agree with the algebra: only II and III hold, so (E) is right.

💡Key takeaway

When a problem uses a strange new symbol, swap in its definition and simplify — two laws survive only if both sides turn into the exact same expression.

Rewrite the operation
Test Law I
Test Law II
Test Law III
Collect the survivors

Rewrite the operation

Test Law I

Test Law II

Test Law III

Collect the survivors

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