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2006 AMC 10B Problem 2

P1-10~1 minPrint
FunctionsAlgebraic ManipulationSet up the equation/formula and compute; no special trick needed

For real numbers xx and yy , define x♠y=(x+y)(x−y)x \spadesuit y = (x+y)(x-y) . What is 3♠(4♠5)3 \spadesuit (4 \spadesuit 5) ?

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Key insight

The spade rule is just the first number squared minus the second number squared. Work out the inside part first, then use its answer, sign included.

Solution

1. Make the rule simpler first. (x+y)(x−y)=x2−y2(x+y)(x-y) = x^2 - y^2, so x♠yx \spadesuit y means "square the first number, square the second, and subtract."
2. Parentheses go first, so start with the inside:
4♠5=42−52=16−25=−9.4 \spadesuit 5 = 4^2 - 5^2 = 16 - 25 = -9.
3. Now the problem is 3♠(−9)3 \spadesuit (-9). The first number is 33 and the second is −9-9:
32−(−9)2=9−81=−72.3^2 - (-9)^2 = 9 - 81 = -72.
Keep the order: it is the first number's square minus the second's, so the answer is negative.

The answer is (A) −72\boxed{\textbf{(A)}\ -72}.

Why this works

A made-up symbol is just a rule with two inputs. Rewriting it in a familiar form (x2−y2x^2 - y^2) makes it quicker to use. When one use of the rule sits inside another, work from the inside out, and carry the inside answer, sign and all, into the outer step.

Alternative approach

Use the rule exactly as written, with no rewriting. Inside: (4+5)(4−5)=9⋅(−1)=−9(4+5)(4-5) = 9 \cdot (-1) = -9. Outside: (3+(−9))(3−(−9))=(−6)(12)=−72(3 + (-9))(3 - (-9)) = (-6)(12) = -72.

Concise solution

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

By the difference of squares, (x+y)(x−y)=x2−y2(x+y)(x-y) = x^2 - y^2, so x♠yx \spadesuit y is simply x2−y2x^2 - y^2.

Work from the inside out. First
4♠5=42−52=16−25=−9.4 \spadesuit 5 = 4^2 - 5^2 = 16 - 25 = -9.
Then
3♠(−9)=32−(−9)2=9−81=−72.3 \spadesuit (-9) = 3^2 - (-9)^2 = 9 - 81 = -72.

The answer is (A) −72\boxed{\textbf{(A)}\ -72}.

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.