AMC 10 · 2006 · #2
Grade 7 arithmeticFor real numbers x and y, define x♠y=(x+y)(x−y). What is 3♠(4♠5)?
Pick an answer.
AMC 10 2006 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 made-up operation $\spadesuit$ is defined for real numbers by $x \spadesuit y = (x+y)(x-y)$. Using this rule, find the value of $3 \spadesuit (4 \spadesuit 5)$.
Givens: The rule $x \spadesuit y = (x+y)(x-y)$ for any real numbers $x$ and $y$; The nested expression $3 \spadesuit (4 \spadesuit 5)$, with the operation inside the parentheses meant to be done first; Answer choices: (A) $-72$, (B) $-27$, (C) $-24$, (D) $24$, (E) $72$
Unknowns: The single number that $3 \spadesuit (4 \spadesuit 5)$ equals
Understand
Restated: A made-up operation $\spadesuit$ is defined for real numbers by $x \spadesuit y = (x+y)(x-y)$. Using this rule, find the value of $3 \spadesuit (4 \spadesuit 5)$.
Givens: The rule $x \spadesuit y = (x+y)(x-y)$ for any real numbers $x$ and $y$; The nested expression $3 \spadesuit (4 \spadesuit 5)$, with the operation inside the parentheses meant to be done first; Answer choices: (A) $-72$, (B) $-27$, (C) $-24$, (D) $24$, (E) $72$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
The expression is one $\spadesuit$ wrapped around another, so Tool #7 (Identify Subproblems) says: solve the inner operation $4 \spadesuit 5$ first, get a single number, then feed that number into the outer $3 \spadesuit (\;)$. Spotting that $(x+y)(x-y)$ is the difference-of-squares pattern $x^2-y^2$ turns each $\spadesuit$ into a quick 'square minus square'. Tool #3 (Eliminate Possibilities) guards the main trap: the operation is not associative, so grabbing $3$ and $4$ first, or swapping which number is $x$, lands on one of the wrong choices.
Execute — Answer: A
6.EE.A.2 Step 1 Read the rule as difference of squares
- The rule $x \spadesuit y = (x+y)(x-y)$ multiplies a sum by a difference.
- Multiplying $(x+y)(x-y)$ out gives $x^2 - y^2$: the cross terms $+xy$ and $-xy$ cancel.
- So every $\spadesuit$ is just 'first number squared minus second number squared'.
- This shortcut is what we will use for both operations.
💡 A sum times its matching difference always collapses to one square minus the other.
6.EE.A.1 Step 2 Do the inner operation first
- Evaluate the operation inside the parentheses, $4 \spadesuit 5$, before anything else.
- Here the left number is $x=4$ and the right number is $y=5$, so it equals $4^2 - 5^2 = 16 - 25$.
- Since $25$ is larger than $16$, the result drops below zero to $-9$.
💡 Squaring the bigger second number makes the difference negative.
7.NS.A.1 Step 3 Feed the result into the outer operation
- Now the problem is $3 \spadesuit (-9)$, with $x=3$ and $y=-9$.
- So it equals $3^2 - (-9)^2 = 9 - 81$.
- Squaring $-9$ gives $+81$ because a negative times a negative is positive, and $9 - 81 = -72$.
- That is choice (A).
- Doing $3 \spadesuit 4$ first instead would give $3^2-4^2=-7$ and miss the parentheses, so the not-associative trap is what separates (A) from the other options.
💡 Take away a much larger square from a small one and you land far below zero.
6.EE.A.2 The rule $x \spadesuit y = (x+y)(x-y)$ multiplies a sum by a difference. Multipl 6.EE.A.1 Evaluate the operation inside the parentheses, $4 \spadesuit 5$, before anything 7.NS.A.1 Now the problem is $3 \spadesuit (-9)$, with $x=3$ and $y=-9$. So it equals $3^2 Review
Reasonableness: Each $\spadesuit$ is 'a square minus a square', and in the final step we subtract $81$ from only $9$, so a large negative number is expected — that immediately rules out the positive choices (D) $24$ and (E) $72$. Squaring $-9$ correctly as $+81$ (not $-81$) is the key move; treating it as $-81$ would give $9-(-81)=90$, which is not even a choice, confirming the sign was handled right. The magnitude $72$ also matches $81-9$, so $-72$ is consistent.
Alternative: Skip the difference-of-squares shortcut and use the product form directly. Inner: $4 \spadesuit 5 = (4+5)(4-5) = (9)(-1) = -9$. Outer: $3 \spadesuit (-9) = (3+(-9))(3-(-9)) = (-6)(12) = -72$. Same answer (A), reached by multiplying a sum times a difference instead of subtracting squares.
CCSS standards used (min grade 7)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Reading the defined operation $x \spadesuit y = (x+y)(x-y)$ and substituting specific numbers for the letters $x$ and $y$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Computing the squares $4^2=16$, $5^2=25$, $3^2=9$, and $(-9)^2=81$ when applying the difference-of-squares form.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Subtracting across zero to get $16-25=-9$ and $9-81=-72$, keeping track of the negative results.)
⭐ When a made-up symbol is defined by a formula, do the operation inside the parentheses first, then plug that number into the outer one — and watch the signs when a square gets subtracted.
⭐ When a made-up symbol is defined by a formula, do the operation inside the parentheses first, then plug that number into the outer one — and watch the signs when a square gets subtracted.
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