2002 AMC 10A Problem 9

P1-10~2 minPrint

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

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?

Show solution without answering

Key insight

Divide both equations by 1001, then add them. The A terms combine to leave exactly A + B + C = 9, so the average is 3.

Solution

We are asked for the average, which only needs the sum $A + B + C$ . So instead of finding each number, look for a way to get the sum directly.

1. Make the numbers smaller. Every number in the equations is a multiple of $1001$ ( $2002 = 2 \times 1001$ , $3003 = 3 \times 1001$ , and so on). Divide each equation by $1001$ :
$C - 2A = 4, \qquad B + 3A = 5.$

2. Add the two equations. On the left, the $A$ parts combine: $-2A + 3A = A$ . So
$A + B + C = 4 + 5 = 9.$
That is exactly the sum we wanted.

Average: $\dfrac{9}{3} = 3$ .

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

Why this works

Two equations cannot pin down three separate numbers, but they can still pin down one combination of them, such as their sum. When a question asks for a sum or an average, try adding or subtracting the given equations before solving for anything. Here the problem was built so that simply adding them works.

Alternative approach

Since the sum is all that matters, pick one number freely and see what happens. Let $C = 0$ . Then $-2A = 4$ gives $A = -2$ , and $B + 3(-2) = 5$ gives $B = 11$ . The sum is $-2 + 11 + 0 = 9$ and the average is $3$ . Any other choice of $C$ gives different numbers but the same sum.

Concise solution

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

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 $\boxed{\textbf{(B)}\ 3}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Intermediate Algebra, Chapter 1: Basic Techniques for Solving Equations (§1.4)
Introduction to Counting & Probability, Chapter 3: Correcting for Overcounting (§3.4)