AMC 10 · 2017 · #25
Easy mode Grade 4A three-digit number "works" if you can rearrange its three digits to make a three-digit multiple of 11 (also between 100 and 999). For example, 121 works because 121 itself is a multiple of 11, and 211 works because its digits can be rearranged into 121. How many of the numbers from 100 to 999 work?
Pick an answer.
AMC 10 2017 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: Count the three-digit numbers from $100$ to $999$ whose digits can be rearranged to form a three-digit multiple of $11$ (also between $100$ and $999$). A number qualifies if at least one arrangement of its digits is such a multiple; for instance $121$ and $211$ both qualify because $121$ is a multiple of $11$.
Givens: We look at every integer from $100$ to $999$ inclusive; A number qualifies if some permutation of its three digits is a multiple of $11$ in the same range; A permutation must itself be a valid three-digit number, so it cannot start with $0$; Answer choices: (A) $226$, (B) $243$, (C) $270$, (D) $469$, (E) $486$
Unknowns: How many three-digit numbers have this property
Understand
Restated: Count the three-digit numbers from $100$ to $999$ whose digits can be rearranged to form a three-digit multiple of $11$ (also between $100$ and $999$). A number qualifies if at least one arrangement of its digits is such a multiple; for instance $121$ and $211$ both qualify because $121$ is a multiple of $11$.
Givens: We look at every integer from $100$ to $999$ inclusive; A number qualifies if some permutation of its three digits is a multiple of $11$ in the same range; A permutation must itself be a valid three-digit number, so it cannot start with $0$; Answer choices: (A) $226$, (B) $243$, (C) $270$, (D) $469$, (E) $486$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems
Checking all $900$ numbers one by one is hopeless, so use Tool #16 (Change Focus): instead of testing each number, start from the multiples of $11$ and count outward to every number that shares a multiple's digits. There are only $81$ such multiples. Tool #2 (Make a Systematic List) then counts, for each multiple, how many valid three-digit rearrangements its digits produce. Tool #7 (Identify Subproblems) sorts the $81$ multiples by how many of their digits repeat, because repeats and zeros change the permutation count, and it flags the double-counting from the reversal rule.
Execute — Answer: A
4.OA.B.4 Step 1 Count the multiples of 11
- List the three-digit multiples of $11$.
- The smallest is $11\times10=110$ and the largest is $11\times90=990$, so they run through $11\times10, 11\times11, \ldots, 11\times90$.
- That is $90-10+1=81$ multiples.
- A number qualifies exactly when its digits can be shuffled into one of these $81$ numbers, so it is enough to count the distinct three-digit numbers reachable by shuffling each multiple's digits.
💡 Counting from the rare multiples outward is far easier than scanning all 900 numbers.
4.OA.A.3 Step 2 A reversal rule that doubles up
- For a number with digits $A$, $B$, $C$, divisibility by $11$ checks $A-B+C$.
- Reversing it to $C$, $B$, $A$ gives $C-B+A$, the same value, so the reverse of any multiple of $11$ is also a multiple of $11$.
- The reverse uses the same three digits, so a multiple and its reverse generate the very same set of rearrangements.
- Whenever a multiple's reverse is a different valid number, those two multiples must not be counted as separate sources, or every shared rearrangement is counted twice.
💡 Flipping a multiple of 11 front-to-back lands on another multiple of 11 with the same digits.
4.OA.A.3 Step 3 Sort the 81 by repeated digits
- Split the $81$ multiples into three groups so each group's permutation count is uniform.
- Group I: all three digits equal, like $aaa=111a$; since $111$ is not a multiple of $11$, there are $0$ of these.
- Group II: exactly two digits equal.
- Group III: all three digits different.
- The remaining $81$ multiples fall into Groups II and III, and within each we separate the ones containing a $0$, because a $0$ cannot lead.
💡 Repeated digits and zeros each shrink how many real numbers a set of digits can spell.
4.NBT.B.5 Step 4 Group II — a repeated digit, no zero
- These are the palindromes $aba$: $121, 242, 363, 484, 616, 737, 858, 979$ — $8$ of them.
- Each has two equal digits and no $0$, so its digits spell $\frac{3!}{2!}=3$ different three-digit numbers, all valid.
- A palindrome is its own reverse, so the reversal rule causes no double-counting here.
💡 Two matching digits cut the six orderings down to three.
4.NBT.B.5 Step 5 Group II — a repeated digit with a zero
- These are $110, 220, 330, 440, 550, 660, 770, 880, 990$ — $9$ of them, each with digits $\{a,a,0\}$.
- The three orderings are $aa0, a0a, 0aa$, but $0aa$ starts with $0$ and is thrown out, leaving $2$ valid numbers each.
- They are not palindromes that pair with another in-range multiple, so no double-counting.
💡 A leading zero is illegal, so a doubled digit plus a zero gives only two real numbers.
4.NBT.B.6 Step 6 Group III — all digits different, with a zero
- Among the $64$ all-different multiples, those containing a $0$ are $209, 308, 407, 506, 605, 704, 803, 902$ — $8$ of them.
- Digits $\{a,b,0\}$ spell $3!=6$ orderings, but the $2$ that start with $0$ are illegal, leaving $4$ valid numbers each.
- The reversal rule pairs them up ($209$ with $902$, etc.), and paired multiples share the same four numbers, so divide by $2$.
💡 Three different digits with a zero make four legal numbers, but reversal counts each pair twice.
4.NBT.B.6 Step 7 Group III — all digits different, no zero
- The rest of Group III are the $64-8=56$ multiples with three different nonzero digits.
- Each spells $3!=6$ valid numbers since no ordering is blocked.
- The reversal rule again pairs each multiple with a distinct in-range partner sharing all six numbers, so divide by $2$.
💡 Three distinct nonzero digits make six numbers, halved because each multiple's reverse is counted too.
3.NBT.A.2 Step 8 Add the groups
- Every qualifying number is counted once across these disjoint cases.
- Add them: $24+18$ from Group II and $16+168$ from Group III.
- That gives $42+184=226$ distinct numbers with the property, which is choice (A).
💡 Separate cases with no overlap simply add up.
4.OA.B.4 List the three-digit multiples of $11$. The smallest is $11\times10=110$ and the 4.OA.A.3 For a number with digits $A$, $B$, $C$, divisibility by $11$ checks $A-B+C$. Rev 4.OA.A.3 Split the $81$ multiples into three groups so each group's permutation count is 4.NBT.B.5 These are the palindromes $aba$: $121, 242, 363, 484, 616, 737, 858, 979$ — $8$ 4.NBT.B.5 These are $110, 220, 330, 440, 550, 660, 770, 880, 990$ — $9$ of them, each with 4.NBT.B.6 Among the $64$ all-different multiples, those containing a $0$ are $209, 308, 40 4.NBT.B.6 The rest of Group III are the $64-8=56$ multiples with three different nonzero d 3.NBT.A.2 Every qualifying number is counted once across these disjoint cases. Add them: $ Review
Reasonableness: The count $226$ is below the $81\times3=243$ you would get by naively giving every multiple three rearrangements, which makes sense because zeros and reversal pairing only remove numbers, never add them. Of the choices, $226$ is the only one under $243$, and every larger choice would require more numbers than the multiples can possibly generate, so (A) is the sole feasible answer. The groups (all-same, one-repeat, all-different) cover the $81$ multiples with no overlap, and within each group the zero and reversal adjustments are accounted for, so nothing is missed or double-counted.
Alternative: Overcount then correct. Give each of the $81$ multiples $6$ orderings and divide by $2$ for the reversal pairing: $81\times3=243$. This over-credits the numbers whose rearrangements include an illegal leading zero. The multiples ending in $0$ contribute $9$ bad orderings ($110,\ldots,990$) and those with a middle-and-last pattern forcing a leading zero contribute $8$ more ($209,\ldots,902$), for $17$ illegal cases. Subtract: $243-17=226$, confirming (A).
CCSS standards used (min grade 4)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing and counting the $81$ three-digit multiples of $11$ that drive the whole count.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Building the multi-step plan: the reversal rule, sorting the multiples into groups, and handling double-counting.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Multiplying each group's size by its number of valid rearrangements, like $8\times3$ and $9\times2$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing by $2$ to remove the reversal double-counting, as in $336\div2=168$.)3.NBT.A.2Fluently add and subtract within 1000 (Adding the group totals $24+18+16+168$ to reach $226$.)
⭐ When few numbers fit a rule, start from those and count outward, adjusting for leading zeros and for shuffles you count twice.
⭐ When few numbers fit a rule, start from those and count outward, adjusting for leading zeros and for shuffles you count twice.
More like this
Same archetype — closest grade level first.