AMC 10 · 2022 · #1
Grade 6 arithmeticProblem
Define x⋄y to be ∣x−y∣ for all real numbers x and y. What is the value of (1⋄(2⋄3))−((1⋄2)⋄3)?
Pick an answer.
AMC 10 2022 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 new symbol is defined: $x \diamond y = |x - y|$, the distance between two numbers. We must compute $\bigl(1 \diamond (2 \diamond 3)\bigr) - \bigl((1 \diamond 2) \diamond 3\bigr)$, where the parentheses tell us which pair to combine first on each side.
Givens: $x \diamond y = |x - y|$ for all real $x, y$; Left side: $1 \diamond (2 \diamond 3)$ — inner pair $(2, 3)$ first; Right side: $(1 \diamond 2) \diamond 3$ — inner pair $(1, 2)$ first; Answer choices: (A) $-2$, (B) $-1$, (C) $0$, (D) $1$, (E) $2$
Unknowns: The numerical value of $\bigl(1 \diamond (2 \diamond 3)\bigr) - \bigl((1 \diamond 2) \diamond 3\bigr)$
Understand
Restated: A new symbol is defined: $x \diamond y = |x - y|$, the distance between two numbers. We must compute $\bigl(1 \diamond (2 \diamond 3)\bigr) - \bigl((1 \diamond 2) \diamond 3\bigr)$, where the parentheses tell us which pair to combine first on each side.
Givens: $x \diamond y = |x - y|$ for all real $x, y$; Left side: $1 \diamond (2 \diamond 3)$ — inner pair $(2, 3)$ first; Right side: $(1 \diamond 2) \diamond 3$ — inner pair $(1, 2)$ first; Answer choices: (A) $-2$, (B) $-1$, (C) $0$, (D) $1$, (E) $2$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
The expression has two separate halves joined by a minus sign, and each half has an inner $\diamond$ that must be done first. Tool #7 (Subproblems) says: compute the left half on its own, compute the right half on its own, then subtract. The order-of-operations rule plus the absolute-value definition do all the work — no algebra needed. Tool #3 (Eliminate) is the multiple-choice safety net: once we have a numerical value, we read off the matching letter.
Execute — Answer: A
6.NS.C.7 Step 1 Compute the inner $\diamond$ on the LEFT side first: $2 \diamond 3 = |2 - 3| = |-1| = 1$.
💡 Absolute value is just the distance between two numbers — Grade 6 "ordering and absolute value" on a number line.
5.OA.A.1 Step 2 Now finish the LEFT side: $1 \diamond (2 \diamond 3) = 1 \diamond 1 = |1 - 1| = 0$.
💡 Substitute the inner result back in, then evaluate — Grade 5 "use parentheses and evaluate numerical expressions".
6.NS.C.7 Step 3 Compute the inner $\diamond$ on the RIGHT side: $1 \diamond 2 = |1 - 2| = |-1| = 1$.
💡 Same distance idea — $1$ and $2$ are $1$ apart.
5.OA.A.1 Step 4 Finish the RIGHT side: $(1 \diamond 2) \diamond 3 = 1 \diamond 3 = |1 - 3| = 2$.
💡 Plug the inner result back in and evaluate the outer $\diamond$ — the parentheses tell us this is the last step.
6.NS.C.5 Step 5 Subtract the two halves: left $-$ right $= 0 - 2 = -2$, which matches choice (A).
💡 Going from $0$ down by $2$ lands at $-2$ — Grade 6 "positive and negative numbers describe quantities".
6.NS.C.7 Compute the inner $\diamond$ on the LEFT side first: $2 \diamond 3 = |2 - 3| = | 5.OA.A.1 Now finish the LEFT side: $1 \diamond (2 \diamond 3) = 1 \diamond 1 = |1 - 1| = 6.NS.C.7 Compute the inner $\diamond$ on the RIGHT side: $1 \diamond 2 = |1 - 2| = |-1| = 5.OA.A.1 Finish the RIGHT side: $(1 \diamond 2) \diamond 3 = 1 \diamond 3 = |1 - 3| = 2$. 6.NS.C.5 Subtract the two halves: left $-$ right $= 0 - 2 = -2$, which matches choice (A) Review
Reasonableness: Sanity check the two halves. Left half closes with $1 \diamond 1$, and any number $\diamond$ itself is $0$ — that's the smallest possible value. Right half closes with $1 \diamond 3$, two numbers that sit $2$ apart, so the value is $2$. A small number minus a bigger number gives a negative result, so $0 - 2 = -2$ has the right sign and size to match choice (A).
Alternative: Tool #1 (Draw a Diagram) on a number line: mark $1, 2, 3$. The left half asks "distance from $1$ to (distance from $2$ to $3$)" $= |1 - 1| = 0$. The right half asks "distance from (distance from $1$ to $2$) to $3$" $= |1 - 3| = 2$. Subtract: $-2$. Same answer, more visual.
CCSS standards used (min grade 6)
5.OA.A.1Use parentheses, brackets, or braces in numerical expressions and evaluate (Honoring the parentheses that tell us which inner $\diamond$ to compute first on each side.)6.NS.C.5Understand that positive and negative numbers describe quantities (Reading the final $0 - 2$ as a negative number, $-2$.)6.NS.C.7Understand ordering and absolute value of rational numbers (Computing each $\diamond$ as the absolute value $|x - y|$, the distance between the two numbers.)
⭐ This AMC 10 problem only needs Grade 6 "absolute value is just the distance between two numbers" — once you read the parentheses carefully, the two halves come out to $0$ and $2$, and the difference is $-2$.
⭐ This AMC 10 problem only needs Grade 6 "absolute value is just the distance between two numbers" — once you read the parentheses carefully, the two halves come out to $0$ and $2$, and the difference is $-2$.
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