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2002 AMC 10A Problem 9

2002 AMC 10A Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Systems of Equations · difficulty 2 of 5 · about 2 min · position P1-10

There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$ , and $1001B + 3003A = 5005$ . What is the average of A, B, and C?

A)
$1$
B)
$3$
C)
$6$
D)
$9$
E)
$\text{Not uniquely determined}$

Answer

Key insight

Add the two equations: the A terms combine to 1001A, so 1001(A+B+C) = 9009 and the average follows without finding A, B, C.

Solution

We need only $A + B + C$ , not the individual values. Every coefficient is a multiple of $1001$ , so divide each equation by $1001$ :
$C - 2A = 4, \qquad B + 3A = 5.$

Adding these two equations makes the $A$ coefficients combine to $-2 + 3 = 1$ :
$A + B + C = 9.$

The average is $\frac{9}{3} = 3$ .

The answer is $\boxed{\textbf{(B)}\ 3}$ .

Why this works

Two equations cannot pin down three unknowns, but they can pin down a particular combination of them, and the problem was built so that the wanted combination $A + B + C$ appears when the equations are added. When a problem asks for a symmetric quantity like a sum or average, look for a linear combination of the givens that produces it directly.

Alternative approach

Since only the sum matters, set $C = 0$ for convenience: then $-2A = 4$ gives $A = -2$ , and $B = 5 - 3(-2) = 11$ . The sum is $-2 + 11 + 0 = 9$ , average $3$ . (Any choice of the free variable gives the same sum.)

The trap

Trying to solve for A, B and C individually and concluding the answer is not uniquely determined because there are only two equations.

Common mistakes

Trying to solve for A, B and C individually and concluding the answer is not uniquely determined because there are only two equations.
Forgetting to divide the total $9$ by $3$ and answering $9$ , choice (D).

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects