AMC 8 · 2022 · #13

Easy mode Grade 4

Problem

Imagine two positive whole numbers. Call them the bigger one and the smaller one.

The bigger one is some amount more than twice the smaller one. We are going to put that "some amount" in the blank. The two numbers also add up to $28$ .

"The bigger number is _____ more than twice the smaller, and the two numbers add up to $28$ ."

How many different positive whole numbers can go in the blank so that the sentence is true?

$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

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

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

#7 Identify Subproblems 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 $3b + k = 28$.

$$a = 2b + k,\;\; a + b = 28 \;\Rightarrow\; 3b + k = 28$$

💡 "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.

#7 Identify Subproblems 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?

$$k = 28 - 3b$$

💡 Turning a two-condition problem into a single subtraction is the Tool #7 "break it down" move — pure Grade 4 multi-step arithmetic.

#6 Guess and Check 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 $9$.

$$28 - 3b \ge 1 \;\Leftrightarrow\; 3b \le 27 \;\Leftrightarrow\; b \le 9$$

💡 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.

#2 Make a Systematic List 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 $b \le 9 < 28$).

$$\begin{array}{c|c|c}b & k = 28-3b & a = 28-b\\\hline 1 & 25 & 27 \\ 2 & 22 & 26 \\ 3 & 19 & 25 \\ 4 & 16 & 24 \\ 5 & 13 & 23 \\ 6 & 10 & 22 \\ 7 & 7 & 21 \\ 8 & 4 & 20 \\ 9 & 1 & 19 \end{array}$$

💡 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.

#2 Make a Systematic List 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).

$$\#\{25, 22, 19, 16, 13, 10, 7, 4, 1\} = 9 \;\Rightarrow\; \textbf{(D)}$$

💡 Counting the entries of a finished list is a Grade 3 two-step word-problem skill — no fancier tools needed.

[1] #7 4.OA.A.2 Translate the two sentences into one arithmetic relationship. "$a$ is $k$ more t

[2] #7 4.OA.A.3 Rearrange to express $k$ as a simple subtraction in terms of $b$. From $3b + k =

[3] #6 4.OA.A.3 Find the largest $b$ that still keeps $k \ge 1$. We need $28 - 3b \ge 1$, i.e. $

[4] #2 4.OA.C.5 Make the systematic list of every valid $(b, k)$ pair, ordered by $b$ from $1$ u

[5] #2 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.8 Solve 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.2 Multiply 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.3 Solve 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.5 Generate 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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