AMC 10 · 2025 · #24
Grade 8 countingCall a positive integer fair if no digit is used more than once, it has no 0s, and no digit is adjacent to two greater digits. For example, 196,23 and 12463 are fair, but 1546,320, and 34321 are not. How many fair positive integers are there?
Pick an answer.
AMC 10 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 positive integer is called fair when all of its digits are different, none of them is $0$, and no digit has two neighbors that are both larger than it. Count how many fair positive integers exist.
Givens: The digits must all be distinct (no digit repeats); No digit may be $0$, so every digit is chosen from $\{1,2,\dots,9\}$; No digit may sit between two strictly larger digits (no interior 'valley'); Examples given as fair: $196,\ 23,\ 12463$; examples given as not fair: $1546,\ 320,\ 34321$; Answer choices: (A) $511$, (B) $2584$, (C) $9841$, (D) $17711$, (E) $19682$
Unknowns: The total count of fair positive integers
Understand
Restated: A positive integer is called fair when all of its digits are different, none of them is $0$, and no digit has two neighbors that are both larger than it. Count how many fair positive integers exist.
Givens: The digits must all be distinct (no digit repeats); No digit may be $0$, so every digit is chosen from $\{1,2,\dots,9\}$; No digit may sit between two strictly larger digits (no interior 'valley'); Examples given as fair: $196,\ 23,\ 12463$; examples given as not fair: $1546,\ 320,\ 34321$; Answer choices: (A) $511$, (B) $2584$, (C) $9841$, (D) $17711$, (E) $19682$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems, #5 Look for a Pattern
This is a 'how many numbers' count over a messy adjacency rule, so Tool #2 (Make a Systematic List) drives it: organize every fair number by which set of digits it uses. Tool #16 (Change Focus) supplies the key reframe — 'no digit between two larger ones' is the same as 'the digits climb to one peak then fall,' turning a checking rule into a shape. Tool #7 (Identify Subproblems) splits the job into 'count the arrangements for one fixed digit set' then 'add over all digit sets.' Tool #5 (Look for a Pattern) recognizes the resulting sum as a binomial expansion so it collapses to a clean $3^9$.
Execute — Answer: C
6.NS.C.7 Step 1 Turn the rule into a shape
- Only an interior digit can be a valley, since the end digits each have just one neighbor.
- If the distinct digits ever went down and then back up, the bottom of that dip would be an interior digit with two larger neighbors — a valley, which is banned.
- So a fair number can never dip in the middle: reading left to right it must strictly increase up to one highest digit (the peak) and then strictly decrease.
- A fair number is exactly a 'mountain' arrangement of distinct nonzero digits.
💡 Forbidding a low point between two highs is the same as saying the digits only go up, then only go down.
7.SP.C.8 Step 2 Count mountains for one digit set
- Fix any set of $k$ distinct digits from $\{1,\dots,9\}$.
- In a mountain the largest chosen digit must be the peak, because nothing beside it may be larger.
- Every other digit then goes on either the up-slope (left of the peak) or the down-slope (right of the peak).
- Once you decide each digit's side, its exact position is forced: the left side must read in increasing order and the right side in decreasing order.
- So building a mountain is just handing each of the remaining $k-1$ digits one of $2$ sides.
💡 Lock the biggest digit at the top, then each leftover digit just flips a two-way switch: left slope or right slope.
7.SP.C.8 Step 3 Add over every digit set
- The number of ways to choose which $k$ digits to use is $\binom{9}{k}$, and each such set yields $2^{k-1}$ fair numbers.
- Letting $k$ run from $1$ to $9$ covers every possible length, so the total is the sum of $\binom{9}{k}\,2^{k-1}$.
- (The $k=1$ term is $\binom{9}{1}\cdot2^{0}=9$, correctly counting the nine one-digit numbers.)
💡 Count each fair number once by first picking its digit set, then arranging that set into a mountain.
8.EE.A.1 Step 4 Collapse the sum and finish
- Pull out the factor $\tfrac12$ so the exponent becomes $2^{k}$, then compare with the binomial theorem $\sum_{k=0}^{9}\binom{9}{k}2^{k}=(1+2)^9=3^9=19683$.
- The full binomial sum includes the missing $k=0$ term $\binom{9}{0}2^{0}=1$, so subtract it before halving.
- That gives $\tfrac{3^9-1}{2}=\tfrac{19682}{2}=9841$, which is choice (C).
- Note $3^9-1=19682$ is the trap choice (E): you land there by forgetting the $\tfrac12$ that comes from fixing the peak.
💡 A sum of $\binom{9}{k}2^{k}$ terms is just $(1+2)^9$ in disguise, so the whole count folds into one power of $3$.
6.NS.C.7 Only an interior digit can be a valley, since the end digits each have just one 7.SP.C.8 Fix any set of $k$ distinct digits from $\{1,\dots,9\}$. In a mountain the large 7.SP.C.8 The number of ways to choose which $k$ digits to use is $\binom{9}{k}$, and each 8.EE.A.1 Pull out the factor $\tfrac12$ so the exponent becomes $2^{k}$, then compare wit Review
Reasonableness: The count must land between two easy anchors. There are $9$ one-digit fair numbers, and using all nine digits gives $2^{8}=256$ mountains, with many sizes in between, so a total in the thousands is sensible — $9841$ fits while tiny (A) $511$ does not. The two large traps confirm the method: $19682=3^9-1$ (choice E) drops the necessary $\tfrac12$, and doubling to $17711$ (choice D) is a Fibonacci lookalike unrelated to this count. Only $9841=\tfrac{3^9-1}{2}$ respects that the peak digit is fixed while the other $k-1$ digits each choose a side.
Alternative: Instead of grouping by digit set, assign each of the digits $1,\dots,9$ to one of three roles independently: 'unused,' 'on the up-slope,' or 'on the down-slope.' That is $3^9$ labelings. The peak is automatically the largest digit that is actually used, so the top-slope choice for that specific largest-used digit is redundant — pairing each valid labeling with its twin shows exactly half are distinct, minus the single all-unused labeling, again giving $\tfrac{3^9-1}{2}=9841$.
CCSS standards used (min grade 8)
6.NS.C.7Understand ordering and absolute value of rational numbers (Reasoning that distinct digits with no interior valley must strictly increase to a single peak and then strictly decrease.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Using the multiplication and combination principles to count $2^{k-1}$ mountains per set and $\binom{9}{k}$ sets of each size.)8.EE.A.1Know and apply the properties of integer exponents (Recognizing $\sum \binom{9}{k}2^{k}=3^9$ and evaluating $3^9=19683$ to finish the count.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Carrying out the final arithmetic $\tfrac{3^9-1}{2}=\tfrac{19682}{2}=9841$.)
⭐ If distinct digits can never dip in the middle, the number must climb to one peak and slide back down — so just fix the biggest digit on top and let every other digit pick a side.
⭐ If distinct digits can never dip in the middle, the number must climb to one peak and slide back down — so just fix the biggest digit on top and let every other digit pick a side.
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