AMC 10 · 2007 · #4
Grade 6 arithmeticThe larger of two consecutive odd integers is three times the smaller. What is their sum?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two odd integers sit next to each other on the number line with no odd integer between them (they are consecutive odd integers). The bigger one equals three times the smaller one. Find the sum of the two integers.
Givens: The two numbers are consecutive odd integers, so the larger is exactly $2$ more than the smaller; The larger integer is three times the smaller integer; Answer choices: (A) $4$, (B) $8$, (C) $12$, (D) $16$, (E) $20$
Unknowns: The two integers themselves, and then their sum
Understand
Restated: Two odd integers sit next to each other on the number line with no odd integer between them (they are consecutive odd integers). The bigger one equals three times the smaller one. Find the sum of the two integers.
Givens: The two numbers are consecutive odd integers, so the larger is exactly $2$ more than the smaller; The larger integer is three times the smaller integer; Answer choices: (A) $4$, (B) $8$, (C) $12$, (D) $16$, (E) $20$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra
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.
Execute — Answer: A
6.EE.B.6 Step 1 Name the two integers
- Let $s$ stand for the smaller odd integer.
- The next consecutive odd integer is always $2$ more, so the larger integer is $s+2$.
- Now both numbers are written with a single unknown.
💡 Odd numbers step by $2$, so the neighbour above $s$ is $s+2$.
6.EE.B.6 Step 2 Turn the words into an equation
- The problem says the larger integer is three times the smaller.
- "Three times the smaller" is $3s$, and the larger integer is $s+2$, so those two amounts are equal: $s+2=3s$.
💡 "Is" means the two sides name the same number, so set them equal.
6.EE.B.7 Step 3 Solve for the smaller integer
- The larger beats the smaller by $2$ (they are consecutive odds).
- It also beats it by $3s-s=2s$, since the larger is $3s$ and the smaller is $s$.
- Those describe the same gap, so $2s=2$, which gives $s=1$.
- Then the larger integer is $s+2=3$.
- Check: $1$ and $3$ are both odd, they are consecutive odd integers, and $3=3\times1$.
💡 The gap between the two numbers is both $2$ and two copies of the smaller, so the smaller must be $1$.
6.EE.A.2 Step 4 Add the two integers
- The integers are $1$ and $3$.
- Their sum is $1+3=4$.
- That matches choice (A).
- The other choices are sums of odd pairs that do not satisfy "larger equals three times smaller": for instance $3+5=8$ gives (B) but $5\neq3\times3$, so those are traps.
💡 Once both integers are known, the sum is just adding them.
6.EE.B.6 Let $s$ stand for the smaller odd integer. The next consecutive odd integer is a 6.EE.B.6 The problem says the larger integer is three times the smaller. "Three times the 6.EE.B.7 The larger beats the smaller by $2$ (they are consecutive odds). It also beats i 6.EE.A.2 The integers are $1$ and $3$. Their sum is $1+3=4$. That matches choice (A). The Review
Reasonableness: 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.
Alternative: Guess and check with small odd numbers. Try $1,3$: is $3=3\times1$? Yes, so the sum is $4$. Try $3,5$: is $5=3\times3=9$? No. Try $5,7$: is $7=15$? No. Only $1,3$ works, confirming (A) without algebra.
CCSS standards used (min grade 6)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Letting $s$ be the smaller integer, writing the larger as $s+2$, and forming the equation $s+2=3s$ from the wording.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Reducing the relationship to $2s=2$ and solving it to get $s=1$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting $s=1$ into $s$ and $s+2$ and adding to get the sum $1+3=4$.)
⭐ Call the smaller number $s$, write the next odd number as $s+2$, turn the sentence into an equation, and the numbers fall out.
⭐ Call the smaller number $s$, write the next odd number as $s+2$, turn the sentence into an equation, and the numbers fall out.
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