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2004 AMC 10A Problem 2

P1-10~2 minPrint
FunctionsSet up the equation/formula and compute; no special trick needed

For any three real numbers aa , bb , and cc , with b≠cb\neq c , the operation ⊗\otimes is defined by: ⊗(a,b,c)=ab−c\otimes(a,b,c)=\frac{a}{b-c} What is ⊗(⊗(1,2,3),⊗(2,3,1),⊗(3,1,2))\otimes(\otimes(1,2,3),\otimes(2,3,1),\otimes(3,1,2)) ?

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Key insight

Work from the inside out. The three inner values are -1, 1 and -3; then use the same rule once more on those three numbers.

Solution

The rule says: take the first number and divide it by (second number minus third number). The inner calculations must be done first, just like parentheses.

1. First inner value: ⊗(1,2,3)=12−3=1−1=−1\otimes(1,2,3) = \dfrac{1}{2-3} = \dfrac{1}{-1} = -1.
2. Second inner value: ⊗(2,3,1)=23−1=22=1\otimes(2,3,1) = \dfrac{2}{3-1} = \dfrac{2}{2} = 1.
3. Third inner value: ⊗(3,1,2)=31−2=3−1=−3\otimes(3,1,2) = \dfrac{3}{1-2} = \dfrac{3}{-1} = -3.
4. Now put these three results, in the same order, into the rule:
⊗(−1, 1, −3)=−11−(−3)=−11+3=−14.\otimes(-1,\,1,\,-3) = \frac{-1}{1-(-3)} = \frac{-1}{1+3} = -\frac{1}{4}.

The answer is (B) −14\boxed{\textbf{(B)}\ -\frac{1}{4}}.

Why this works

A made-up operation is just a recipe. When one recipe is used inside another, finish the inner ones first, the same way you handle nested parentheses. Writing each middle result on its own line makes the minus signs much easier to keep straight.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Work from the inside out. The rule divides the first argument by the difference of the other two:

  • ⊗(1,2,3)=12−3=−1\otimes(1,2,3) = \dfrac{1}{2-3} = -1,
  • ⊗(2,3,1)=23−1=1\otimes(2,3,1) = \dfrac{2}{3-1} = 1,
  • ⊗(3,1,2)=31−2=−3\otimes(3,1,2) = \dfrac{3}{1-2} = -3.

These three values become the new arguments, in that order:
⊗(−1, 1, −3)=−11−(−3)=−14.\otimes(-1,\,1,\,-3) = \frac{-1}{1-(-3)} = \frac{-1}{4}.

The answer is (B) −14\boxed{\textbf{(B)}\ -\frac{1}{4}}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.