2007 AMC 10A Problem 4

P1-10~1 minPrint

Linear Equations & Word Problems Number Properties Set up the equation/formula and compute; no special trick needed

The larger of two consecutive odd integers is three times the smaller. What is their sum?

Show solution without answering

Key insight

Two odd numbers in a row are 2 apart. If the bigger one is three times the smaller, the difference of 2 must equal twice the smaller.

Solution

1. Set up. Call the smaller odd number $x$ . The next odd number is $x + 2$ (for example $5$ and $7$ ).
2. Use the condition. The larger is three times the smaller:
$x + 2 = 3x.$
3. Solve. Take $x$ from both sides: $2 = 2x$ , so $x = 1$ .
4. Check. The numbers are $1$ and $3$ . Both are odd, they are next to each other, and $3 = 3 \times 1$ .
5. Sum. $1 + 3 = 4$ .

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

Why this works

"Consecutive odd" or "consecutive even" numbers are always $2$ apart, so one letter is enough to name both. Checking at the end that your numbers really are odd catches a wrong setup.

Alternative approach

Just try the smallest pairs of odd numbers: $1$ and $3$ . Is $3$ three times $1$ ? Yes, so you are done: the sum is $4$ . (Larger pairs cannot work: for $3$ and $5$ , three times $3$ is already $9$ , far past $5$ .)

Concise solution

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

Let the smaller odd integer be $x$ ; the next odd integer is $x + 2$ . The condition says
$x + 2 = 3x \quad\Longrightarrow\quad 2x = 2 \quad\Longrightarrow\quad x = 1.$
The integers are $1$ and $3$ ; indeed $3 = 3 \cdot 1$ . Their sum is $1 + 3 = 4$ .

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

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 7: Proportion (§7.4)
Introduction to Algebra, Chapter 20: Special Functions (§20.3)