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2005 AMC 10A Problem 2

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2005 AMC 10A Problem 2AMC 10 Step by Step
Name:                             Date: September 13, 2026
Functions · difficulty 1 of 5 · about 1 min · position P1-10

For each pair of real numbers a≠ba \neq b , define the operation ⋆\star as

(a⋆b)=a+ba−b.(a \star b) = \frac{a+b}{a-b}.

What is the value of ((1⋆2)⋆3)((1 \star 2) \star 3) ?

  1. A)

    −23-\frac{2}{3}

  2. B)

    −15-\frac{1}{5}

  3. C)

    00

  4. D)

    12\frac{1}{2}

  5. E)

    This value is not defined.\text{This value is not defined.}

Answer

C

Key insight

Evaluate the inner operation first: 1 star 2 equals -3, and then -3 star 3 has numerator zero.

Solution

Work from the inside out. First,
1⋆2=1+21−2=3−1=−3.1 \star 2 = \frac{1 + 2}{1 - 2} = \frac{3}{-1} = -3.
Next, substitute this result as the left input:
(−3)⋆3=−3+3−3−3=0−6=0.(-3) \star 3 = \frac{-3 + 3}{-3 - 3} = \frac{0}{-6} = 0.
The denominator is nonzero at every step, so the value is defined and equals 00.

The answer is (C) 0\boxed{\textbf{(C)}\ 0}.

Why this works

A made-up operation is just a two-variable function; nested parentheses tell you which evaluation to do first. The only way such an expression can be "not defined" is a zero denominator, which the condition a≠ba \neq b rules out here at both steps.

The trap

Seeing a zero somewhere and answering 'not defined', when only a zero denominator (a = b) would make the value undefined.

Common mistakes

  • Seeing a zero somewhere and answering 'not defined', when only a zero denominator (a = b) would make the value undefined.
  • Evaluating 1⋆(2⋆3)1 \star (2 \star 3) instead of (1⋆2)⋆3(1 \star 2) \star 3; the operation is not associative, and the wrong grouping gives 1+(−5)1−(−5)=−23\frac{1 + (-5)}{1 - (-5)} = -\frac{2}{3}, choice (A).

Techniques

Set up the equation/formula and compute; no special trick needed