AMC 10 · 2007 · #4

Grade 6 algebra

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Direct Ratio Scale-Up

linear-equations-one-varparity convert-to-algebra ↑ Prerequisites: linear-equations-one-var

📏 Short solution 💡 1 insight

Problem

Two integers are consecutive odd integers, so the larger is exactly 2 more than the smaller. The larger integer is three times the smaller integer. What is the sum of the two integers?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Let s be the smaller odd integer. The next odd integer up is 2 bigger, so the larger is s+2 — one unknown covers both.

smaller=s, larger=s+2

2STEP 2

Turn the words into an equation

"Three times the smaller" is 3s, and the larger is s+2. Those name the same integer, so s+2=3s.

s+2=3s

3STEP 3

Solve for the smaller integer

The gap is 2 and also 3s-s=2s, so 2s=2 and s=1. The larger is s+2=3; indeed 3=3×1.

2s=2 → s=1, s+2=3

4STEP 4

Add the two integers

The integers are 1 and 3, so the sum is 1+3=4 — choice (A). Traps like 3+5=8 fail 5≠3×3.

1+3=4 → (A)

Answer

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

Name the two integers

Turn the words into an equation

Solve for the smaller integer

Add the two integers

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