AMC 10 · 2025 · #13
Grade 7 counting
Pick an answer.
The question is a 'how many ways' count, so a systematic list is the right engine. To keep the list short, first draw the ring and label the positions, then solve an easier version by fixing the color of one sector and multiplying at the end. The remaining count breaks into a few clean cases based on where the second same-colored sector sits.
Label the ring positions
Number the slots and write who touches whom.
Naming the spots turns a vague 'circle' into a fixed list you can fill in one position at a time.
7.SP.C.8Draw A DiagramFix sector 1, multiply by 3 later
Fix slot one and multiply by three later.
Locking one color in place removes repeated work, and the missing colors are recovered by one clean times-3 at the end.
Locking one colour in place removes repeated work, and the missing cases come back with one multiplication.
▸ Why?
Turning the ring carries one arrangement onto another without stretching, so those copies are the same picture.
▸ Why?
Each arrangement matches the same fixed number of copies, so the count is corrected by one factor.
Where can the second red go?
The second red can only sit at slots 3, 4, or 5.
The one forbidden move, red next to red, cuts the choices down to a short list of positions.
7.SP.C.8Identify SubproblemsCount each case by listing
Listing the cases gives 8.
Once red is pinned down, the rest is a tiny alternating pattern you can just write out and count.
7.SP.C.8Make A Systematic ListMultiply by 3 for all top colors
Times three is 24.
Three interchangeable colors mean three copies of the same count, so one multiplication finishes the job.
3.OA.A.1Make A Systematic ListPin one color in place, count the few ways the rest can fit, then multiply back by the colors you set aside.
- Label the ring positions
- Fix sector 1, multiply by 3 later
- Where can the second red go?
- Count each case by listing
- Multiply by 3 for all top colors
More like this
Same archetype — closest grade level first.
