AMC 8 · 2022 · #13
Grade 4 algebranumber-theoryProblem
How many positive integers can fill the blank in the sentence below?
“One positive integer is _____ more than twice another, and the sum of the two numbers is .”
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: We have two positive whole numbers. The first one equals twice the second one, plus some extra positive whole number $k$ (the value in the blank). The two numbers also add up to $28$. The question is: how many different positive whole numbers $k$ can go in the blank?
Givens: Two positive integers, call them $a$ (the bigger one) and $b$ (the other one); $a$ is $k$ more than twice $b$, where $k$ is also a positive integer; $a + b = 28$; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Unknowns: The number of different positive integer values of $k$ that make the sentence true
Understand
Restated: We have two positive whole numbers. The first one equals twice the second one, plus some extra positive whole number $k$ (the value in the blank). The two numbers also add up to $28$. The question is: how many different positive whole numbers $k$ can go in the blank?
Givens: Two positive integers, call them $a$ (the bigger one) and $b$ (the other one); $a$ is $k$ more than twice $b$, where $k$ is also a positive integer; $a + b = 28$; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #6 Guess and Check, #7 Identify Subproblems
The question "how many values of $k$ work?" begs for a Tool #2 systematic list — pick an ordering for $b$ (the smaller of the two numbers) and list every $b$ that produces a valid positive $k$. To set up the list cleanly we first use Tool #7 (Subproblems): combine the two facts ($a = 2b + k$ and $a + b = 28$) into a single arithmetic relationship $3b + k = 28$, so $k = 28 - 3b$. Then Tool #6 (Guess and Check) lets us walk $b = 1, 2, 3, \dots$ and stop the moment $k$ would drop below $1$. This avoids reaching for Tool #13 (algebra) when ordinary subtraction and a list do the job.
Execute — Answer: D
4.OA.A.2 Step 1 - Translate the two sentences into one arithmetic relationship.
- "$a$ is $k$ more than twice $b$" means $a = 2b + k$.
- Combine that with $a + b = 28$ by substituting: $(2b + k) + b = 28$, which simplifies to $3b + k = 28$.
💡 "Twice another" is a multiplicative comparison — a Grade 4 word-problem move — and we just add the second number to both sides of the count.
4.OA.A.3 Step 2 - Rearrange to express $k$ as a simple subtraction in terms of $b$.
- From $3b + k = 28$ we get $k = 28 - 3b$.
- Now the question becomes: for which positive integers $b$ is $28 - 3b$ also a positive integer?
💡 Turning a two-condition problem into a single subtraction is the Tool #7 "break it down" move — pure Grade 4 multi-step arithmetic.
4.OA.A.3 Step 3 - Find the largest $b$ that still keeps $k \ge 1$.
- We need $28 - 3b \ge 1$, i.e.
- $3b \le 27$, i.e.
- $b \le 9$.
- So $b$ can be any whole number from $1$ to $9$.
💡 Asking "how big can $b$ be before $k$ runs out?" is Tool #6 (Guess and Check) on the boundary; the test reduces to $27 \div 3 = 9$, a Grade 4 fact.
4.OA.C.5 Step 4 - Make the systematic list of every valid $(b, k)$ pair, ordered by $b$ from $1$ up to $9$.
- Check along the way that the bigger number $a = 28 - b$ is also a positive integer (it always is, since $b \le 9 < 28$).
💡 Generating the table by the rule "subtract $3$ from $k$ every time $b$ goes up by $1$" is exactly the Grade 4 pattern-from-a-rule standard.
3.OA.D.8 Step 5 - Count the distinct positive-integer values of $k$ in the list: $25, 22, 19, 16, 13, 10, 7, 4, 1$ — nine different numbers.
- Each $b$ produces a different $k$ (because the rule $k = 28 - 3b$ is strictly decreasing), so no duplicates.
- The answer is $9$, which is choice (D).
💡 Counting the entries of a finished list is a Grade 3 two-step word-problem skill — no fancier tools needed.
4.OA.A.2 Translate the two sentences into one arithmetic relationship. "$a$ is $k$ more t 4.OA.A.3 Rearrange to express $k$ as a simple subtraction in terms of $b$. From $3b + k = 4.OA.A.3 Find the largest $b$ that still keeps $k \ge 1$. We need $28 - 3b \ge 1$, i.e. $ 4.OA.C.5 Make the systematic list of every valid $(b, k)$ pair, ordered by $b$ from $1$ u 3.OA.D.8 Count the distinct positive-integer values of $k$ in the list: $25, 22, 19, 16, Review
Reasonableness: Sanity-check the two extreme rows of the table. When $b = 1$: the bigger number is $a = 27$, and $2b + k = 2 + 25 = 27$. Yes — $a + b = 28$ and $a$ is $25$ more than twice $b$. When $b = 9$: $a = 19$, and $2b + k = 18 + 1 = 19$. Yes — $a + b = 28$ and $a$ is just $1$ more than twice $b$. Trying $b = 10$ would force $k = 28 - 30 = -2$, which is not a positive integer, so the list really does stop at $b = 9$. Nine values is consistent, and choice (D) $9$ is the answer.
Alternative: Tool #3 (Eliminate Possibilities) on the multiple-choice list: the candidate counts are $6, 7, 8, 9, 10$. Since $k$ must be a positive integer of the form $28 - 3b$, the largest valid $k$ is at $b = 1$ giving $k = 25$, and the smallest valid $k$ is at $b = 9$ giving $k = 1$. The valid $k$ values $\{1, 4, 7, 10, 13, 16, 19, 22, 25\}$ are exactly the positive integers $\le 25$ that leave remainder $1$ when divided by $3$ — and there are $9$ of those, eliminating every choice except (D).
CCSS standards used (min grade 4)
3.OA.D.8Solve two-step word problems using four operations within 100 (Counting the finished list of valid $k$ values and assembling the final answer from the listed entries.)4.OA.A.2Multiply or divide to solve word problems involving multiplicative comparison (Interpreting the phrase "$k$ more than twice another" as the multiplicative-comparison expression $a = 2b + k$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining the two given facts into $3b + k = 28$ and bounding $b$ via $3b \le 27$ so $b \le 9$.)4.OA.C.5Generate a number or shape pattern following a given rule (Building the table of $(b, k)$ pairs by the rule "each time $b$ increases by $1$, $k$ decreases by $3$".)
⭐ This AMC 8 problem only needs Grade 4 "twice as many" thinking and a careful list you already know how to make!
⭐ This AMC 8 problem only needs Grade 4 "twice as many" thinking and a careful list you already know how to make!