AMC 8 · 2025 · #13

Easy mode Grade 4

Problem

Picture the list of even numbers $2, 4, 6, \ldots, 50$ .

Take each one and divide it by $7$ . Write down the remainder you get each time.

Some remainders will come up more often than others. Which of the histograms below shows how many times each remainder (from $0$ up to $6$ ) appeared?

Pick an answer.

(A)

(histogram) bar heights for remainders 0–6: 3, 4, 4, 3, 4, 3, 4

(B)

(histogram) bar heights for remainders 0–6: 3, 4, 4, 4, 3, 3, 4

(C)

(histogram) bar heights for remainders 0–6: 3, 4, 4, 4, 4, 3, 3

(D)

(histogram) bar heights for remainders 0–6: 4, 3, 4, 3, 4, 3, 4

(E)

(histogram) bar heights for remainders 0–6: 4, 4, 3, 4, 3, 4, 3

AMC 8 2025 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Take the $25$ even numbers $2, 4, 6, \ldots, 50$ and divide each one by $7$. Record the remainder ($0$ through $6$) every time. Among the five histograms, find the one whose bar heights match how many times each remainder shows up.

Givens: The numbers under consideration are the even integers $2, 4, 6, \ldots, 50$; Each number is divided by $7$ and the remainder is recorded; Possible remainders are $0, 1, 2, 3, 4, 5, 6$; Five candidate histograms (A)-(E) list bar heights for remainders $0$-$6$

Unknowns: How many times each remainder $0, 1, 2, 3, 4, 5, 6$ occurs, and which histogram matches that frequency table

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

Dividing $2, 4, 6, \ldots$ by $7$ produces a repeating cycle of remainders, so Tool #5 (Look for a Pattern) is the natural primary tool — compute the first several remainders, spot the cycle, then count efficiently. Tool #2 (Systematic List) supports this by listing the first seven remainders in order so the cycle is visible. Tool #3 (Eliminate Possibilities) is the multiple-choice closer: once we have the seven counts, only one of (A)-(E) matches and the rest can be crossed out.

Execute — Answer: A

#2 Make a Systematic List 4.OA.A.3 Step 1

Count how many even numbers are in the list.
The even numbers from $2$ to $50$ are $2 \cdot 1, 2 \cdot 2, \ldots, 2 \cdot 25$, so there are exactly $25$ of them.
The seven bar heights must therefore add to $25$.

$$25 = 2 \cdot 1, 2 \cdot 2, \ldots, 2 \cdot 25 \;\Rightarrow\; 25 \text{ numbers}$$

💡 Pairing each even number with a counting number $1$-$25$ is a Grade 4 multi-step word-problem move.

#2 Make a Systematic List 4.NBT.B.6 Step 2

List the remainders of the first several even numbers when divided by $7$ to expose the cycle.
Doing this carefully shows a length-$7$ pattern: $2, 4, 6, 1, 3, 5, 0$.

$$2 \div 7 \to 2,\; 4 \div 7 \to 4,\; 6 \div 7 \to 6,\; 8 \div 7 \to 1,\; 10 \div 7 \to 3,\; 12 \div 7 \to 5,\; 14 \div 7 \to 0$$

💡 Finding whole-number quotients and remainders is exactly the Grade 4 division-with-remainder skill.

#5 Look for a Pattern 4.OA.C.5 Step 3

Confirm the cycle repeats.
The next number $16$ gives $16 \div 7 = 2$ remainder $2$, matching the start of the cycle.
Adding $14$ ($= 2 \cdot 7$) to any number keeps the same remainder mod $7$, so the $7$-step block $(2, 4, 6, 1, 3, 5, 0)$ repeats forever.

$$16 \div 7 \to 2 \;\; (\text{same as } 2 \div 7)$$

💡 Spotting a repeating cycle in a generated number list is the Grade 4 pattern-rule standard in action.

#5 Look for a Pattern 4.NBT.B.6 Step 4

Count how many complete cycles fit in $25$ numbers.
Divide: $25 \div 7 = 3$ with remainder $4$.
That means the full cycle $(2, 4, 6, 1, 3, 5, 0)$ runs $3$ times (covering $21$ numbers), and then the first $4$ entries of one more cycle appear.

$$25 = 7 \cdot 3 + 4$$

💡 Splitting $25$ into $3$ full groups of $7$ plus $4$ leftovers is the same Grade 4 division-with-remainder idea, used at the cycle level.

#2 Make a Systematic List 4.OA.A.3 Step 5

Tally each remainder.
After $3$ complete cycles, every remainder $0$-$6$ has been hit $3$ times.
The $4$ leftover numbers ($44, 46, 48, 50$) follow the start of the cycle, giving remainders $2, 4, 6, 1$ — so add $1$ to each of those bins.

$$44 \div 7 \to 2,\; 46 \div 7 \to 4,\; 48 \div 7 \to 6,\; 50 \div 7 \to 1$$

💡 Combining "$3$ from each cycle" with "$+1$ for the leftovers" is a Grade 4 two-step word-problem combination.

#3 Eliminate Possibilities 3.MD.B.3 Step 6

Write the final bar heights in remainder order $0, 1, 2, 3, 4, 5, 6$ and check the total.
Rem $0$: $3$.
Rem $1$: $3+1=4$.
Rem $2$: $3+1=4$.
Rem $3$: $3$.
Rem $4$: $3+1=4$.
Rem $5$: $3$.
Rem $6$: $3+1=4$.
Sum $= 3+4+4+3+4+3+4 = 25$.

$$[3, 4, 4, 3, 4, 3, 4],\quad 3+4+4+3+4+3+4 = 25 \;\checkmark$$

💡 Reading bar heights off a frequency tally is the Grade 3 scaled-bar-graph skill.

#3 Eliminate Possibilities 3.MD.B.3 Step 7

Match the heights to the five histograms and eliminate the wrong ones.
Only (A) shows $[3, 4, 4, 3, 4, 3, 4]$.
(B) is $[3, 4, 4, 4, 3, 3, 4]$, (C) is $[3, 4, 4, 4, 4, 3, 3]$, (D) is $[4, 3, 4, 3, 4, 3, 4]$, (E) is $[4, 4, 3, 4, 3, 4, 3]$ — none match.

$$\text{(A)} = [3, 4, 4, 3, 4, 3, 4] \;\Rightarrow\; \textbf{(A)}$$

💡 Comparing our computed frequencies to each histogram is a Grade 3 bar-graph interpretation task.

[1] #2 4.OA.A.3 Count how many even numbers are in the list. The even numbers from $2$ to $50$ a

[2] #2 4.NBT.B.6 List the remainders of the first several even numbers when divided by $7$ to exp

[3] #5 4.OA.C.5 Confirm the cycle repeats. The next number $16$ gives $16 \div 7 = 2$ remainder

[4] #5 4.NBT.B.6 Count how many complete cycles fit in $25$ numbers. Divide: $25 \div 7 = 3$ with

[5] #2 4.OA.A.3 Tally each remainder. After $3$ complete cycles, every remainder $0$-$6$ has bee

[6] #3 3.MD.B.3 Write the final bar heights in remainder order $0, 1, 2, 3, 4, 5, 6$ and check t

[7] #3 3.MD.B.3 Match the heights to the five histograms and eliminate the wrong ones. Only (A)

Review

Reasonableness: There are $25$ even numbers and $7$ possible remainders, so on average each remainder should land $25 / 7 \approx 3.57$ times. With whole-number counts, we'd expect a mix of $3$s and $4$s — exactly what (A) shows ($3$ threes and $4$ fours). The bar heights sum to $25$ as required, and the $4$ "extra" remainders ($1, 2, 4, 6$) are exactly the remainders of the four leftover numbers $50, 44, 46, 48$, so the count is internally consistent.

Alternative: Tool #16 (Change Focus / Complement): instead of tracking every even number, note that $2k \bmod 7$ for $k = 1, 2, \ldots, 25$ is the same as $2k \bmod 7$ where $k$ runs through residues mod $7$. Each residue class for $k$ mod $7$ contains either $3$ or $4$ values of $k$ in $\{1, \ldots, 25\}$ (since $25 = 7 \cdot 3 + 4$), and the residues $k \equiv 1, 2, 3, 4 \pmod 7$ each appear $4$ times — which by doubling map to remainders $2, 4, 6, 1$. So those four remainders get $4$ each and the rest get $3$ each, matching (A).

CCSS standards used (min grade 4)

3.MD.B.3 Draw and interpret scaled picture graphs and bar graphs (Reading bar heights off a histogram and matching the computed frequencies $[3, 4, 4, 3, 4, 3, 4]$ to the correct choice (A).)
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Dividing each even number (and the count $25$) by $7$ to record the quotient and remainder.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Recognizing that the remainders of $2, 4, 6, 8, \ldots$ divided by $7$ form a repeating length-$7$ cycle $(2, 4, 6, 1, 3, 5, 0)$.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Combining "$3$ from each of $3$ full cycles" with "$+1$ for each of the $4$ leftover remainders" to tally the seven bar heights.)

⭐ This AMC 8 problem only needs Grade 4 division-with-remainder and pattern-spotting you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 4)

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