AMC 10 · 2005 · #2
Grade 7 arithmeticFor each pair of real numbers a=b, define the operation ⋆ as
(a⋆b)=a−ba+b.
What is the value of ((1⋆2)⋆3)?
Pick an answer.
AMC 10 2005 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 operation $\star$ is defined for real numbers $a\neq b$ by $(a\star b)=\dfrac{a+b}{a-b}$. Working from the inside out, evaluate $((1\star 2)\star 3)$, and decide whether the result is defined.
Givens: The rule is $(a\star b)=\dfrac{a+b}{a-b}$; The rule applies only when $a\neq b$; The expression to evaluate is $((1\star 2)\star 3)$; Answer choices: (A) $-\frac{2}{3}$, (B) $-\frac{1}{5}$, (C) $0$, (D) $\frac{1}{2}$, (E) the value is not defined
Unknowns: The numerical value of $((1\star 2)\star 3)$, or whether it is undefined
Understand
Restated: A new operation $\star$ is defined for real numbers $a\neq b$ by $(a\star b)=\dfrac{a+b}{a-b}$. Working from the inside out, evaluate $((1\star 2)\star 3)$, and decide whether the result is defined.
Givens: The rule is $(a\star b)=\dfrac{a+b}{a-b}$; The rule applies only when $a\neq b$; The expression to evaluate is $((1\star 2)\star 3)$; Answer choices: (A) $-\frac{2}{3}$, (B) $-\frac{1}{5}$, (C) $0$, (D) $\frac{1}{2}$, (E) the value is not defined
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
The expression is one operation nested inside another, so Tool #7 (Identify Subproblems) splits it into two clean pieces: first compute the inner value $1\star 2$, then feed that result into $(\cdot)\star 3$. Tool #3 (Eliminate Possibilities) handles the only trap: choice (E) says the value is not defined, which happens only if a denominator $a-b$ equals $0$; checking each denominator confirms whether (E) can be dismissed.
Execute — Answer: C
6.EE.A.2 Step 1 Plug into the inner operation
- Apply the rule to the inside first.
- For $1\star 2$ take $a=1$ and $b=2$, so $a+b=1+2=3$ and $a-b=1-2=-1$.
- Since $1\neq 2$, the denominator is nonzero and the operation is allowed here.
💡 A defined operation is just a formula: substitute the two numbers into the top and bottom exactly as written.
7.NS.A.2 Step 2 Finish the inner value
- Divide to finish the inner result: a positive divided by a negative is negative, so $\dfrac{3}{-1}=-3$.
- The inner expression $1\star 2$ equals $-3$, which now becomes the left input of the outer operation.
💡 Dividing by $-1$ just flips the sign, turning $3$ into $-3$.
7.NS.A.1 Step 3 Evaluate the outer operation
- Now compute $(-3)\star 3$ with $a=-3$ and $b=3$.
- The numerator is $a+b=-3+3=0$ and the denominator is $a-b=-3-3=-6$.
- Since $-3\neq 3$, the denominator $-6$ is not zero, so the value is defined and choice (E) is ruled out.
- A fraction with a zero top and a nonzero bottom equals $0$, so $\dfrac{0}{-6}=0$.
- The value of $((1\star 2)\star 3)$ is $0$, choice (C).
💡 A fraction is $0$ exactly when its top is $0$ and its bottom is not, so the sign of the bottom does not matter.
6.EE.A.2 Apply the rule to the inside first. For $1\star 2$ take $a=1$ and $b=2$, so $a+b 7.NS.A.2 Divide to finish the inner result: a positive divided by a negative is negative, 7.NS.A.1 Now compute $(-3)\star 3$ with $a=-3$ and $b=3$. The numerator is $a+b=-3+3=0$ a Review
Reasonableness: Both denominators were checked to be nonzero ($1-2=-1$ and $-3-3=-6$), so the operation was legal at every step and choice (E) is correctly eliminated. The final answer hinges on the outer numerator $-3+3$ being $0$; adding a number to its opposite always gives $0$, which is a reliable way to land on exactly $0$. Among the choices only (C) is $0$, and the other numeric choices would each require a nonzero numerator, which the computation does not produce.
Alternative: Skip the outer denominator entirely. Once the inner value is $-3$, the outer numerator is $(-3)+3=0$. A fraction equals $0$ as soon as its numerator is $0$ and its denominator is nonzero, so after a one-line check that $-3-3=-6\neq 0$, the answer is $0$ without dividing.
CCSS standards used (min grade 7)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Treating $(a\star b)=\frac{a+b}{a-b}$ as a formula and substituting $a=1,\,b=2$ to evaluate the inner operation.)7.NS.A.2Apply and extend previous understandings of multiplication and division to divide rational numbers (Dividing $3$ by $-1$ to get the negative inner value $-3$.)7.NS.A.1Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers (Computing the outer numerator $-3+3=0$ and denominator $-3-3=-6$ with negative numbers.)
⭐ When an operation is spelled out as a formula, plug the numbers in innermost first, and remember a fraction is $0$ exactly when its top is $0$ and its bottom is not.
⭐ When an operation is spelled out as a formula, plug the numbers in innermost first, and remember a fraction is $0$ exactly when its top is $0$ and its bottom is not.
More like this
Same archetype — closest grade level first.