AMC 10 · 2013 · #22
Grade 7 countingThe regular octagon ABCDEFGH has its center at J. Each of the vertices and the center are to be associated with one of the digits 1 through 9, with each digit used once, in such a way that the sums of the numbers on the lines AJE, BJF, CJG, and DJH are all equal. In how many ways can this be done?
Pick an answer.
AMC 10 2013 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 regular octagon ABCDEFGH has center J. The four lines AJE, BJF, CJG, DJH are the diameters, each passing through two opposite vertices and the center. Place the digits 1 through 9, each used exactly once, on the eight vertices and the center so that the three-number sum on each of the four lines is the same. Count how many placements work.
Givens: There are 9 spots: 8 vertices and the center J; The digits 1 through 9 are each used exactly once; The four lines are AE, BF, CG, DH; each line holds two opposite vertices plus the center J; The three numbers on each line must add to the same total
Unknowns: The number of valid ways to place the digits
Understand
Restated: A regular octagon ABCDEFGH has center J. The four lines AJE, BJF, CJG, DJH are the diameters, each passing through two opposite vertices and the center. Place the digits 1 through 9, each used exactly once, on the eight vertices and the center so that the three-number sum on each of the four lines is the same. Count how many placements work.
Givens: There are 9 spots: 8 vertices and the center J; The digits 1 through 9 are each used exactly once; The four lines are AE, BF, CG, DH; each line holds two opposite vertices plus the center J; The three numbers on each line must add to the same total
Plan
Primary tool: #2 Make a Systematic List
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
This is a 'how many ways' counting question, so the plan is to pin down the structure, then multiply independent choices. First name the common line sum with a variable to get an equation linking it to the center digit; that equation limits the center to just a few digits. For each allowed center digit, the leftover eight digits pair up into equal-sum pairs; then I count how many ways those pairs can be placed on the four lines and flipped end for end. Multiplying the independent counts gives the total.
Execute — Answer: C
6.EE.B.6 Step 1 Add all four line sums
- Let S be the common sum on each line.
- Add the four line sums together.
- Each vertex sits on exactly one line, so all eight vertices are counted once; the center J sits on all four lines, so it is counted four times.
- The eight vertices plus the center use every digit 1 through 9, which total 45.
- So four times S equals 45 plus three extra copies of the center digit J.
💡 Adding the lines counts the shared center four times, so the center digit is the only thing that breaks the plain total of 45.
4.OA.B.4 Step 2 Find which center digits work
- Since 4S is a multiple of 4, the quantity 45 + 3J must be divisible by 4.
- Test the digits 1 through 9: 45 + 3J is a multiple of 4 only when J is 1, 5, or 9 (giving 48, 60, and 72).
- Every other digit fails.
- So the center must be one of these 3 digits.
💡 Only a center digit that makes the grand total a multiple of 4 can split evenly into four equal line sums.
4.OA.A.3 Step 3 Pair the leftover eight digits
- For a fixed center J, each line's two vertices must add to S minus J, the same target for every line.
- That target is (45 - J)/4: for J=1 it is 11, for J=5 it is 10, for J=9 it is 9.
- In each case the remaining eight digits split into four pairs with that sum in exactly one way.
- For example, with J=1 the pairs are (2,9), (3,8), (4,7), (5,6).
💡 The eight leftovers are forced into one tidy set of equal-sum pairs, so there is nothing to choose at this stage.
7.SP.C.8 Step 4 Place and flip the four pairs
- Now count placements for one center digit.
- The four pairs can be matched to the four lines AE, BF, CG, DH in 4!
- = 24 ways.
- On each line the pair's two digits can go on the two opposite vertices in 2 orders, and the four lines flip independently, giving 2 to the fourth power = 16.
- Multiply: 24 times 16 = 384 placements for each allowed center digit.
💡 Choosing which pair goes on which line and which end each digit takes are independent, so their counts multiply.
7.SP.C.8 Step 5 Combine the three center choices
- There are 3 valid center digits, and each gives 384 placements, with no overlap between the cases since they use different center digits.
- Add or, equivalently, multiply the 3 cases by 384 to get the grand total.
- This gives 3 times 384 = 1152, so the answer is (C).
💡 Each center digit runs its own independent count, so total the three equal cases.
6.EE.B.6 Let S be the common sum on each line. Add the four line sums together. Each vert 4.OA.B.4 Since 4S is a multiple of 4, the quantity 45 + 3J must be divisible by 4. Test t 4.OA.A.3 For a fixed center J, each line's two vertices must add to S minus J, the same t 7.SP.C.8 Now count placements for one center digit. The four pairs can be matched to the 7.SP.C.8 There are 3 valid center digits, and each gives 384 placements, with no overlap Review
Reasonableness: The total 1152 factors as 3 x 24 x 16, matching the three independent stages: 3 center choices, 4! ways to assign pairs to lines, and 2^4 ways to flip pairs end for end. It sits sensibly between the choices: 384 (C's answer for a single center) is exactly one center's count, so forgetting to multiply by 3 would land on (A); 1152 is 384 tripled, confirming (C). The value is comfortably under the total 9! arrangements, as expected once the equal-sum constraint prunes most placements.
Alternative: Instead of pairing first, build a placement slot by slot going clockwise. Pick the center (3 ways), then fill vertex A (8 ways). Vertex A fixes its opposite E, since together they must hit the target sum. Vertex B is then any of the 6 remaining, which fixes F; then vertex C is one of the 4 left, fixing G; then D is one of the 2 left, fixing H. That gives 3 x 8 x 6 x 4 x 2 = 1152, the same answer.
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Naming the common line sum S and deriving 4S = 45 + 3J by adding the four line sums)4.OA.B.4Determine whether a whole number is a multiple of a given number (Testing which center digits make 45 + 3J a multiple of 4, leaving J = 1, 5, or 9)4.OA.A.3Solve multistep problems with the four operations and assess reasonableness (Finding the fixed pair-sum target (45 - J)/4 and the unique set of four equal-sum pairs)7.SP.C.8Find outcome counts of compound events using organized lists and the counting principle (Multiplying 4! ways to assign pairs to lines, 2^4 ways to flip them, and the 3 center choices)
⭐ The center digit sits on every line, so only 1, 5, or 9 can balance the four sums; after that the leftovers pair up in one way and you just count how to place and flip them, giving 3 x 24 x 16 = 1152.
⭐ The center digit sits on every line, so only 1, 5, or 9 can balance the four sums; after that the leftovers pair up in one way and you just count how to place and flip them, giving 3 x 24 x 16 = 1152.
More like this
Same archetype — closest grade level first.
