2025 AMC 8 Problem 22
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To help you master 2025 AMC 8 Problem 22, this page includes:
- The original AMC problem to try, plus the correct answer
- Clear, step-by-step solutions using different methods
- Key ideas, useful formulas, and takeaways you can apply to similar problems
- Related problems from past AMC exams, plus new practice problems that build the same skills
Want more practice? Scroll down to download Think Academy’s AMC 8 practice packs and keep building your AMC 8 skills.
2025 AMC8 Problem 22
22.
A classroom has a row of 35 coat hooks. Paulina likes coats to be equally spaced, so that there is the same number of empty hooks before the first coat, after the last coat, and between every coat and the next one. Suppose there is at least 1 coat and at least 1 empty hook. How many different numbers of coats can satisfy Paulina’s pattern?
Solution
Factoring by Grouping Method
Express the total number of hooks algebraically, then add a constant to both sides so the equation can be factored into the product of two integers.
Simon’s Favorite Factoring Trick (Factoring by Grouping)
Step 1: Define the variables and identify the number of gaps
Let be the number of coats and be the number of empty hooks in each equally spaced gap. To find the total number of gaps, we count the gap before the first coat, the gaps between the coats, and the gap after the last coat. For coats, there are gaps between them, plus the gaps at the ends.
Step 2: Set up the equation for the total number of hooks
The total number of hooks on the rack is the sum of the hooks holding coats and the empty hooks. We are given that the total is .
Step 3: Factor the equation
To solve this equation with two integer variables, we want to express it as a product of factors. Notice the term . If we add to both sides of our equation, we can group the terms to create a common factor of .
Step 4: Establish minimum bounds based on the problem conditions
The problem states there is at least coat, meaning . It also states there is at least empty hook overall. Since all gaps must have the same number of empty hooks, each gap must have at least empty hook, meaning . We apply these inequalities to our factors.
Step 5: Find the valid factors of 36
We need to find all integer factor pairs of where both factors are at least . The full list of factors of is . We exclude and because neither factor is allowed to be . The remaining factors represent the valid values for .
There are 7 valid values for the number of coats (Choice D). As a quick check, if , then coats. This means , so empty hooks per gap. Total hooks = coats + gaps empty hooks = , which works perfectly.
Algebraic Division Shortcut
Instead of factoring, solve the total hook equation for the number of empty hooks, and use divisibility rules to see when the result is an integer.
Step 1: Solve the hook equation for the empty hooks
Using the same initial equation for the total number of hooks, we can isolate (the number of empty hooks per gap). Since must be an integer, this turns the problem into a divisibility check.
Step 2: Rewrite the fraction to reveal the factors
To easily see when is an integer, we can manipulate the numerator so that it includes a multiple of the denominator . We subtract and add in the numerator.
Step 3: Determine valid values for the number of coats
For to be an integer, must divide perfectly into . Just as before, requires . Also, requires that the fraction evaluate to at least , meaning . The divisors of strictly between and give us exactly valid values for .
There are 7 values. This confirms the result from Method 1 without needing to spot the grouping trick, relying purely on fraction division.
Phantom Object Method (Periodic Grouping)
By adding one imaginary coat to the right end of the rack, the entire arrangement becomes a repeating sequence of identical blocks, turning the problem into simple division.
Step 1: Visualize the repeating pattern of hooks
Let be the number of empty hooks in each gap. The rack starts with empty hooks, followed by coat. This pattern repeats for every coat. However, after the final coat, there is one last gap of empty hooks that does not have a coat after it.
Step 2: Introduce a phantom coat to complete the pattern
Imagine placing one extra, imaginary ‘phantom’ coat at the very right end of the rack. Now, every single gap of empty hooks is perfectly paired with a coat immediately to its right. The entire rack is made of identical repeating blocks.
Step 3: Relate the blocks to the total number of items
If there are real coats, adding the phantom coat means there are identical blocks in total. The total number of hooks, including our phantom one, is . The total items must equal the number of blocks multiplied by the size of each block.
Step 4: Count the valid configurations
Because there is at least one real coat, the number of blocks must be at least . Because there is at least one empty hook total, the block size must also be at least . We are looking for all integer factors of that are between and inclusive.
There are 7 valid numbers of blocks, and subtracting the phantom coat gives 7 valid numbers of actual coats. Matches Choice D.
Similar Problems in Real Exams
The following five AMC 8 problems are closely related to AMC 8 2025 Question 22. These real AMC 8 problems test the same key concept—Factors, Divisors & Divisor Functions. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.
2022 AMC 8 Problem 3
2020 AMC 8 Problem 17
2018 AMC 8 Problem 18
2015 AMC 8 Problem 22
2009 AMC 8 Problem 11
Adapted Problems to Practice
What Formulas Do You Need for This Problem?
- Fencepost Principle (Number of Intervals):
If you have objects arranged in a line, there are gaps strictly between them. If you also count the space before the first object and after the last object, there are total gaps.
Applies when arranging identical items in a sequence and evaluating the spaces separating them. We needed to know how many gaps of empty hooks existed relative to the number of coats to write our main equation.
- Simon's Favorite Factoring Trick (SFFT):
An equation of the form can be factored by adding to both sides, resulting in .
and are variables (usually integers), and is a constant. Our total hook equation was , which perfectly matches this structure. Adding allowed us to factor it into .
Key Takeaway: Remember This Strategy for Next Time
A repeating sequence of items that almost forms identical blocks, but falls short by exactly one item at the end.
Try: Imagine adding a 'phantom' object to complete the final block.
Why: It makes the total length perfectly divisible by the block size, reducing a messy algebraic equation into a simple multiplication and factoring problem.
An equation with two integer variables where variables are multiplied and added, like .
Try: Add to both sides of the equation to factor it by grouping.
Why: Adding creates the terms needed to factor out common binomials, transforming addition into multiplication so you can just list integer factors.
Phrases like 'at least 1' or 'at most' in a problem requesting a count of valid integers.
Try: Immediately translate these phrases into mathematical inequalities (e.g., ) and apply them to your final factor pairs.
Why: Mathematical equations often have extreme solutions (like ) that work algebraically but physically violate the story's conditions. Setting bounds prevents you from double-counting these invalid cases.
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