AMC 10 · 2007 · #3

Grade 6 algebra
linear-equations-one-varparity convert-to-algebra ↑ Prerequisites: linear-equations-one-var
📏 Short solution 💡 1 insight
Problem
Two consecutive odd integers are such that the bigger is three times the smaller. Find their sum.

Pick an answer.

(A)
4
(B)
8
(C)
12
(D)
16
(E)
20
How to solve
Strategy Introduce a Variable

The smaller integer is the one unknown that controls everything else, so Tool #4 (Introduce a Variable) names it s and writes the larger one as s+2 (the next odd number is always 2 bigger). Tool #13 (Convert to Algebra) then turns the sentence "the larger is three times the smaller" into an equation, which pins down s exactly instead of guessing.

1STEP 1

Name the two integers

Consecutive odd numbers differ by two.

smaller=s, larger=s+2
2STEP 2

Turn the words into an equation

The sentence becomes a one-step equation.

s+2=3s
3STEP 3

Solve for the smaller integer

Solving gives the smaller as 1.

2s=2 → s=1, s+2=3
4STEP 4

Add the two integers

Adding gives 4, choice (A).

1+3=4 → (A)
Answer
4
Test the pair against every condition: 1 and 3 are odd, they are consecutive odd integers (differ by 2), and the larger 3 is exactly three times the smaller 1. All three hold, and their sum 4 is answer (A). It is also the smallest choice, which fits: tripling forces the smaller number to be tiny, so the sum stays small. Any larger smaller-value, say 3, would need a larger of 9, but 9 is not 2 more than 3, so no bigger pair works.
💡Key takeaway

Call the smaller number s, write the next odd number as s+2, turn the sentence into an equation, and the numbers fall out.

  • Name the two integers
  • Turn the words into an equation
  • Solve for the smaller integer
  • Add the two integers