AMC 8 · 2025 · #22
Easy mode Grade 4Problem
Picture a long row of 35 coat hooks on a wall, side by side.
Paulina hangs some coats on the hooks. She likes the coats to be spread out evenly. That means the number of empty hooks before the very first coat, the number of empty hooks after the very last coat, and the number of empty hooks between any two coats that are next to each other are all the same.
There must be at least 1 coat hanging up. There must also be at least 1 empty hook somewhere.
How many different numbers of coats are possible?
Pick an answer.
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.
Toolkit + CCSS Solution
Understand
Restated: A row of $35$ coat hooks must be filled so the empty hooks are arranged in equal groups: one group before the first coat, one group after the last coat, and one equal group between every pair of adjacent coats. Both the number of coats and the size of each empty-hook group must be at least $1$. How many different values for the number of coats are possible?
Givens: Total hooks in the row $= 35$; Each coat sits on a single hook; The empty hooks are split into equal groups before the first coat, after the last coat, and between every two neighboring coats; At least $1$ coat and at least $1$ empty hook in every group; Answer choices: (A) $2$, (B) $4$, (C) $5$, (D) $7$, (E) $9$
Unknowns: How many different values of $c$ (the number of coats) make Paulina's pattern fit exactly $35$ hooks
Understand
Restated: A row of $35$ coat hooks must be filled so the empty hooks are arranged in equal groups: one group before the first coat, one group after the last coat, and one equal group between every pair of adjacent coats. Both the number of coats and the size of each empty-hook group must be at least $1$. How many different values for the number of coats are possible?
Givens: Total hooks in the row $= 35$; Each coat sits on a single hook; The empty hooks are split into equal groups before the first coat, after the last coat, and between every two neighboring coats; At least $1$ coat and at least $1$ empty hook in every group; Answer choices: (A) $2$, (B) $4$, (C) $5$, (D) $7$, (E) $9$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #6 Guess and Check, #5 Look for a Pattern
The question asks "how many different numbers of coats" — a perfect trigger for Tool #2 (Systematic List). For each possible coat count $c$, the empty hooks left over are $35 - c$, and they must split evenly into $c+1$ equal groups. So for each $c$ we just check whether $c+1$ divides $35 - c$ evenly — pure Guess & Check (Tool #6) on a small, finite list. As we walk the list we notice a clean divisibility pattern (Tool #5): the leftover $35 - c$ equals $36 - (c+1)$, so $(c+1)$ divides $35 - c$ exactly when $(c+1)$ divides $36$. That insight turns the search into "which divisors of $36$ are $\ge 2$?" — no algebra needed.
Execute — Answer: D
3.OA.B.6 Step 1 - Set up the picture.
- If $c$ coats are placed equally, the $35 - c$ empty hooks must split evenly into $c+1$ groups (one before, one between every adjacent pair, one after).
- So the gap size $k$ is $k = \dfrac{35 - c}{c + 1}$, and the requirement is that $k$ comes out as a whole number $\ge 1$.
💡 "Equal groups" of empty hooks is the same Grade 3 idea as splitting a number of cookies evenly onto a number of plates.
4.OA.A.3 Step 2 - Walk the list of possible $c$ values from small to large, computing $(35 - c) \div (c + 1)$ each time.
- The biggest $c$ can be is $17$ (because $c$ and at least $c+1$ empty hooks together must fit in $35$, so $c + (c+1) \le 35$).
💡 Listing the candidates in order is the Tool #2 move and a standard Grade 4 multi-step word-problem habit.
4.OA.B.4 Step 3 - Carry out the check on each $c$.
- A value works when the division gives a whole number $\ge 1$.
💡 Each check is just "does this number divide evenly?" — exactly Grade 4 factor/multiple thinking.
4.OA.B.4 Step 4 - Notice the hidden pattern that explains why $c = 1, 2, 3, 5, 8, 11, 17$ are the winners.
- Rewrite the numerator: $35 - c = 36 - (c + 1)$.
- So $\dfrac{35 - c}{c + 1} = \dfrac{36}{c+1} - 1$, which is a whole number exactly when $c + 1$ is a divisor of $36$.
- The divisors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18, 36$, and we need $c + 1 \ge 2$ (since $c \ge 1$) and also $c \ge 1$ rules out $c + 1 = 36$ unless $k \ge 1$ (check: $c+1=36 \Rightarrow c=35$, leaves $0$ empty hooks, fails).
- So $c + 1 \in \{2, 3, 4, 6, 9, 12, 18\}$.
💡 Spotting that $36$ controls the whole problem turns the search into "list the factor pairs of $36$," which is a Grade 4 standard skill.
4.OA.A.3 Step 5 - Count the surviving values of $c$.
- From $c + 1 \in \{2, 3, 4, 6, 9, 12, 18\}$ we get $c \in \{1, 2, 3, 5, 8, 11, 17\}$ — that is $7$ different possibilities, matching choice (D).
💡 Counting the items left after eliminating bad cases is the final "how many ways?" answer.
3.OA.B.6 Set up the picture. If $c$ coats are placed equally, the $35 - c$ empty hooks mu 4.OA.A.3 Walk the list of possible $c$ values from small to large, computing $(35 - c) \d 4.OA.B.4 Carry out the check on each $c$. A value works when the division gives a whole n 4.OA.B.4 Notice the hidden pattern that explains why $c = 1, 2, 3, 5, 8, 11, 17$ are the 4.OA.A.3 Count the surviving values of $c$. From $c + 1 \in \{2, 3, 4, 6, 9, 12, 18\}$ we Review
Reasonableness: Sanity-check the two extremes. With $c = 1$ coat, the gap size is $k = 17$: pattern is $17$ empty + $1$ coat + $17$ empty = $35$ hooks. With $c = 17$ coats, $k = 1$: pattern is $1$-empty-$1$-coat repeated, total $1 + 17 \cdot 2 = 35$. Both check out exactly, so the answer set $\{1, 2, 3, 5, 8, 11, 17\}$ is real and complete. Also $7$ is choice (D) and isn't the trap value $9$ (= number of divisors of $36$ before we throw out the $1$-and-$36$ edge cases) or $5$ (the symmetric solution count if we mistakenly only counted up to $c = 5$). The two edge-case eliminations exactly account for the difference $9 - 2 = 7$.
Alternative: Tool #13 (Convert to Algebra) gives Simon's Favorite Factoring Trick: $c + k(c+1) = 35 \Rightarrow (k+1)(c+1) = 36$, then count factor pairs of $36$ with both factors $\ge 2$. That's the slicker high-school path, but our Tool #2 + #5 route reaches the same $36$ insight using only Grade-4 divisibility — no algebraic manipulation needed.
CCSS standards used (min grade 4)
3.OA.B.6Understand division as an unknown-factor problem (Setting up the gap size $k = (35 - c) / (c + 1)$ as "split the leftover hooks evenly into $c+1$ equal groups," the Grade 3 equal-shares meaning of division.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Walking the systematic list of coat counts and counting how many surviving values remain — the multi-step whole-number computation behind the final tally.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Checking which $c+1$ values divide $35 - c$ evenly, and noticing that this is the same as listing the factor pairs of $36$ (and keeping only those with both factors $\ge 2$).)
⭐ This AMC 8 problem only needs the Grade 4 "factor pairs of 36" skill you already know — no algebra required!
⭐ This AMC 8 problem only needs the Grade 4 "factor pairs of 36" skill you already know — no algebra required!
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