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2002 AMC 10B Problem 2

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2002 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 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

D is just a recipe: multiply the three inputs, divide by their sum, and 48/12 = 4.

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 $\boxed{\textbf{(C)}\ 4}$ .

Why this works

A newly defined symbol like $D(a,b,c)$ is nothing more than a formula with three slots. Fill the slots, then simplify. The "nonzero" hypothesis only guarantees the denominator is defined and plays no role in the arithmetic.

The trap

Mixing up which operation goes where (adding on top, multiplying below) and getting 12/48 = 1/4 or a wrong integer.

Common mistakes

Mixing up which operation goes where (adding on top, multiplying below) and getting 12/48 = 1/4 or a wrong integer.
Reporting the numerator $48$ or a partial product such as $24$ (choice E) instead of finishing the division.

Techniques

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