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2006 AMC 10B Problem 2

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2006 AMC 10B Problem 2AMC 10 Step by Step

Name: Date: September 13, 2026

Functions · difficulty 1 of 5 · about 1 min · position P1-10

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

A)
$-72$
B)
$-27$
C)
$-24$
D)
$24$
E)
$72$

Answer

Key insight

The operation is just x^2 - y^2, so evaluate the inner one first: 4 spade 5 = -9, then 3 spade (-9) = 9 - 81.

Solution

By the difference of squares, $(x+y)(x-y) = x^2 - y^2$ , so $x \spadesuit y$ is simply $x^2 - y^2$ .

Work from the inside out. First
$4 \spadesuit 5 = 4^2 - 5^2 = 16 - 25 = -9.$
Then
$3 \spadesuit (-9) = 3^2 - (-9)^2 = 9 - 81 = -72.$

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

Why this works

A made-up operation is just a formula with two inputs; recognizing it as $x^2 - y^2$ removes the multiplication step. Nested operations are evaluated innermost first, exactly like nested function calls, and the sign of the inner result must be carried faithfully into the outer one.

The trap

Dropping the sign of the inner result (using 9 instead of -9), or squaring -9 and forgetting the square is positive.

Common mistakes

Dropping the sign of the inner result (using 9 instead of -9), or squaring -9 and forgetting the square is positive.
Computing $(3 \spadesuit 4) \spadesuit 5$ by ignoring the parentheses.

Techniques

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