AMC 10 · 2014 · #24
Grade 7 geometry-2dThe numbers 1,2,3,4,5 are to be arranged in a circle. An arrangement is bad if it is not true that for every n from 1 to 15 one can find a subset of the numbers that appear consecutively on the circle that sum to n. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?
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Pick an answer.
AMC 10 2014 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: Place $1,2,3,4,5$ around a circle. For a target $n$ from $1$ to $15$, you are allowed to add up numbers that sit next to each other along the circle (a single number counts, and so does the whole circle). Call the arrangement $\textit{bad}$ if at least one target $n$ in $1$ to $15$ can never be formed this way. Arrangements that match after a rotation or a mirror flip count as one. Count the bad arrangements.
Givens: The five numbers $1,2,3,4,5$ sit in a circle; their total is $1+2+3+4+5 = 15$; A usable sum comes from a block of numbers that are consecutive along the circle (an arc); An arc may be a single number, or the whole circle, or anything in between; Rotations and reflections of the same picture are considered identical; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: How many distinct circular arrangements are bad (fail to reach some $n$ between $1$ and $15$)
Understand
Restated: Place $1,2,3,4,5$ around a circle. For a target $n$ from $1$ to $15$, you are allowed to add up numbers that sit next to each other along the circle (a single number counts, and so does the whole circle). Call the arrangement $\textit{bad}$ if at least one target $n$ in $1$ to $15$ can never be formed this way. Arrangements that match after a rotation or a mirror flip count as one. Count the bad arrangements.
Givens: The five numbers $1,2,3,4,5$ sit in a circle; their total is $1+2+3+4+5 = 15$; A usable sum comes from a block of numbers that are consecutive along the circle (an arc); An arc may be a single number, or the whole circle, or anything in between; Rotations and reflections of the same picture are considered identical; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #16 Change Focus / Count the Complement, #1 Draw a Diagram, #3 Eliminate Possibilities
"How many arrangements" with a finite pool screams Tool #2 (Make a Systematic List) — but a raw list of $120$ orderings is wasteful. First shrink the work two ways. Tool #16 (Count the Complement) notices that the total is $15$, so an arc summing to $k$ leaves a complementary arc summing to $15-k$; that instantly makes $1$--$5$, $10$--$15$ free and pairs the doubtful targets as $6\leftrightarrow 9$ and $7\leftrightarrow 8$, so we only ever test $6$ and $7$. Tool #1 (Draw a Diagram) with the rotation-and-reflection rule cuts $120$ orderings down to just $12$ distinct circles. Then Tool #2 lists those $12$ and Tool #3 (Eliminate) crosses off every circle that manages to build both $6$ and $7$, leaving the bad ones.
Execute — Answer: B
2.NBT.B.5 Step 1 List the sums you always get
- The five numbers total $15$.
- Single numbers hand you $1,2,3,4,5$.
- An arc of four numbers is the whole circle minus one number, so it equals $15-1,15-2,15-3,15-4,15-5 = 14,13,12,11,10$, giving you $10,11,12,13,14$.
- The full circle gives $15$.
- None of this depends on the order.
- So every arrangement, good or bad, can already build $1,2,3,4,5,10,11,12,13,14,15$.
💡 Singles, all-but-one, and the whole circle are automatic, so most targets can never be the problem.
4.OA.A.3 Step 2 Only 6 and 7 can go wrong
- The only targets not yet guaranteed are $6,7,8,9$.
- Use the complement: if some arc sums to $6$, the rest of the circle is also an arc and sums to $15-6=9$; so being able to make $6$ is the same as being able to make $9$.
- Likewise making $7$ is the same as making $8$ (since $15-7=8$).
- Therefore an arrangement is bad exactly when it fails to build $6$, or fails to build $7$.
- We only have to test those two.
💡 An arc and everything left over always add to $15$, so reaching a number automatically reaches its partner.
4.G.A.3 Step 3 Only 12 different circles exist
- There are $5! = 120$ ways to write the numbers in order, but rotations and reflections collapse them.
- Pin the number $1$ at the top: that removes the $5$ rotations.
- Each circle also equals its own mirror image, which removes a factor of $2$.
- So the number of genuinely different circles is $120 \div (5\times 2) = 12$.
- These twelve are the only cases to check.
💡 Spinning or flipping a circle does not change which numbers are neighbors, so those copies count once.
7.SP.C.8 Step 4 Test each circle for 6 and 7
- You can build $6$ as $1{+}5$, as $2{+}4$, or as $1{+}2{+}3$; you can build $7$ as $2{+}5$, as $3{+}4$, or as $1{+}2{+}4$.
- So a circle makes $6$ unless $1\&5$ are apart, $2\&4$ are apart, and $1,2,3$ are not three-in-a-row; similarly for $7$.
- Running through all $12$ circles, ten of them manage both $6$ and $7$ and are good.
- Exactly two fail.
- Circle $1\text{-}2\text{-}5\text{-}3\text{-}4$ has neighbor pairs $\{1,2\},\{2,5\},\{5,3\},\{3,4\},\{4,1\}$ and no three-in-a-row equal to $1,2,3$, so it can never make $6$ (its reachable sums are $1,2,3,4,5,7,8,10,11,12,13,14,15$ — missing $6$ and $9$).
- Circle $1\text{-}3\text{-}2\text{-}4\text{-}5$ can never make $7$ (missing $7$ and $8$).
💡 Since $6$ and $7$ each form only three ways, checking whether any of those pieces sits together is quick for every circle.
7.SP.C.8 Step 5 Count the bad circles
- Two of the twelve circles are bad: $1\text{-}2\text{-}5\text{-}3\text{-}4$ (no way to make $6$) and $1\text{-}3\text{-}2\text{-}4\text{-}5$ (no way to make $7$).
- They are genuinely different circles — in one, $5$ sits between $2$ and $3$; in the other, $5$ sits between $4$ and $1$ — so neither is a rotation or reflection of the other.
- The count of bad arrangements is $2$, which is choice (B).
💡 Blocking $6$ forces one circle and blocking $7$ forces a different one, and there is no third way.
2.NBT.B.5 The five numbers total $15$. Single numbers hand you $1,2,3,4,5$. An arc of four 4.OA.A.3 The only targets not yet guaranteed are $6,7,8,9$. Use the complement: if some a 4.G.A.3 There are $5! = 120$ ways to write the numbers in order, but rotations and refle 7.SP.C.8 You can build $6$ as $1{+}5$, as $2{+}4$, or as $1{+}2{+}3$; you can build $7$ a 7.SP.C.8 Two of the twelve circles are bad: $1\text{-}2\text{-}5\text{-}3\text{-}4$ (no w Review
Reasonableness: Both bad circles were verified by listing their complete set of arc sums, not just spot-checked. For $1\text{-}2\text{-}5\text{-}3\text{-}4$: neighbor sums are $3,7,8,7,5$ and three-in-a-row sums are $8,10,12,8,7$; together with $1$--$5$, $10$--$14$, $15$ that is $\{1,2,3,4,5,7,8,10,11,12,13,14,15\}$ — genuinely missing $6$ and $9$. For $1\text{-}3\text{-}2\text{-}4\text{-}5$ the same accounting misses $7$ and $8$. A full computer-style sweep of all $12$ circles returns exactly these two, so the answer $2$ is not an undercount from a missed case, and it lands squarely inside the choice list at (B).
Alternative: Focus on the neighbors of $5$ (Tool #16). Since we only need to block $6$ or block $7$: to block $6$ you must keep $1\&5$ apart, keep $2\&4$ apart, and keep $1,2,3$ from being three-in-a-row — walking those requirements around the circle forces exactly $1\text{-}2\text{-}5\text{-}3\text{-}4$. To block $7$ the mirror-image requirements force exactly $1\text{-}3\text{-}2\text{-}4\text{-}5$. Two forced circles, so the answer is $2$ without listing all twelve.
CCSS standards used (min grade 7)
2.NBT.B.5Fluently add and subtract within 100 (Adding the numbers along each arc, and computing $15-k$ to see that singles, all-but-one arcs, and the whole circle cover $1$--$5$ and $10$--$15$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Using the complement relation $k \leftrightarrow 15-k$ to pair $6$ with $9$ and $7$ with $8$, reducing the whole problem to testing just $6$ and $7$.)4.G.A.3Recognize a line of symmetry for a two-dimensional figure (Treating a circle and its mirror flip (and its rotations) as the same arrangement, cutting $120$ orderings down to $12$ distinct circles.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Systematically enumerating the $12$ distinct circles and eliminating the ten that build both $6$ and $7$, leaving the two bad arrangements.)
⭐ The whole circle is $15$, so every arc and its leftover add to $15$ — that makes all targets free except $6$ and $7$, and only two of the twelve different circles fail to build one of them.
⭐ The whole circle is $15$, so every arc and its leftover add to $15$ — that makes all targets free except $6$ and $7$, and only two of the twelve different circles fail to build one of them.
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