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2002 AMC 10B Problem 2

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2002 AMC 10B Problem 2AMC 10 Step by Step

Name: Date: October 28, 2026

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

For the nonzero numbers a, b, and c, define

$D(a,b,c)=\frac{abc}{a+b+c}$

Find $D(2,4,6)$ .

A) $1$
B) $2$
C) $4$
D) $6$
E) $24$

Answer

Key insight

The rule D just says: multiply the three numbers, add the three numbers, and divide the product by the sum.

Solution

The letters $a$ , $b$ , $c$ are just slots. Put $2$ , $4$ , $6$ into them.

Top: multiply the three numbers, $2 \cdot 4 \cdot 6 = 48$ .
Bottom: add the three numbers, $2 + 4 + 6 = 12$ .
Divide: $\dfrac{48}{12} = 4$ .

The answer is $\boxed{\textbf{(C)}\ 4}$ .

Why this works

A made-up function like $D(a,b,c)$ is only a recipe. Follow the recipe exactly, keeping track of which part goes on top and which goes on the bottom. The word "nonzero" is there only so the rule always makes sense; it does not change the arithmetic.

The trap

Swapping the top and bottom, putting the sum over the product, and getting 12/48 = 1/4.

Concise solution

Substitute $a = 2$ , $b = 4$ , $c = 6$ into the definition. The numerator is the product $2\cdot 4\cdot 6 = 48$ and the denominator is the sum $2 + 4 + 6 = 12$ , so
$D(2,4,6) = \frac{48}{12} = 4.$

The answer is $\boxed{\textbf{(C)}\ 4}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 16: Functions
Introduction to Algebra, Chapter 17: Graphing Functions (§17.2)
Intermediate Algebra, Chapter 19: Functional Equations (§19.1, §19.2)

Techniques

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