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2007 AMC 10A Problem 4

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2007 AMC 10A Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 1 min · position P1-10

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

A)
$4$
B)
$8$
C)
$12$
D)
$16$
E)
$20$

Answer

Key insight

Consecutive odd integers differ by 2, so write them as x and x + 2 and solve x + 2 = 3x.

Solution

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

Why this works

"Consecutive odd" (or "consecutive even") integers are spaced $2$ apart, so one variable suffices. The tripling condition then becomes a one-step linear equation. Checking that the solution really is odd guards against a setup error.

Alternative approach

Odd integers $x$ and $3x$ that differ by $2$ means $2x = 2$ , so $x = 1$ at a glance; or simply test the smallest odd pairs: $(1, 3)$ works immediately.

The trap

Writing consecutive odd integers as x and x + 1, which gives a non-integer solution.

Common mistakes

Writing consecutive odd integers as $x$ and $x + 1$ , which gives a non-integer solution.
Reporting the larger integer ( $3$ ) or the smaller one instead of their sum.

Techniques

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