Competition · AMC preparation · step 4 of 4
AMC 8 · 2017 · #15
Grade 4 counting
Pick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem is purely spatial — we are walking on a grid — so Tool #1 (Draw a Diagram) is the natural starting point. With the picture in hand we use Tool #7 (Identify Subproblems) to split the 3-move walk into three independent counting questions: A → M, then M → C, then C → 8. Tool #5 (Look for a Pattern) helps once we notice that every M has the same number of valid C-neighbors and every C has the same number of valid 8-neighbors, so we can multiply the three small counts instead of enumerating all paths.
Count moves from A to M
The central A's four orthogonal neighbors are all M's (up, down, left, right), so the first move has 4 choices.
Just describing positions like "above, below, left, right" of A is a Kindergarten geometry idea.
K.G.A.1Draw A DiagramCount moves from M to C
From any M the neighbors are the already-used A plus three C's, so by symmetry every M offers 3 C-choices.
Spotting that the same count of 3 repeats at every M is a Grade 4 "find the repeating rule" pattern observation.
4.OA.C.5Look For A PatternCount moves from C to 8
Both kinds of C — the inner corner and the outer tip — touch exactly two 8's, so each C gives 2 choices for the last move.
Confirming that the count of 2 holds for both types of C is another Grade 4 pattern-rule check.
4.OA.C.5Look For A PatternMultiply the three counts
Multiply the three independent counts by the multiplication principle: 4 × 3 × 2 = 24 paths — answer choice (D).
Multiplying the choices at each step is exactly the Grade 3 "number of groups times number per group" multiplication idea.
The number of different AMC8 paths equals the count of first-move choices times the count of second-move choices times the count of third-move choices.
▸ Why?
A path is fixed by three moves made in order — pick a cell for M, then one for C, then one for 8 — and no two different sequences of picks trace the same path and no path comes from two sequences, so counting paths is the same as counting these sequences of picks.
▸ Why?
At each stage the number of next cells to choose does not depend on which cells were picked earlier: the center borders the same four M cells, every M borders the same number of new C cells, and every C borders the same number of 8 cells, so the choice at each stage is independent of the earlier ones.
▸ Why?
The whole arrangement looks identical after a quarter turn or a mirror flip, so any M can be turned or flipped onto any other M with its neighbors carried along; matching cells must therefore have matching neighbor counts, giving every M the same number of C neighbors and every C the same number of 8 neighbors.
▸ Why?
Because each stage offers a fixed number of options no matter what was chosen before, the three stages are independent choices, and independent choices combine by multiplying their counts.
This AMC 8 problem only needs Grade 4 pattern-spotting plus the simple Grade 3 idea that the total number of paths is the product of the choices at each step!
- Count moves from A to M
- Count moves from M to C
- Count moves from C to 8
- Multiply the three counts
A parent dashboard for the family lives at sensimlab.com.
More like this
Same problem type. Tags show what each one shares with this problem.
- AMC 8 2010 #25 Gr 4Similar levelSame sub-type: Constrained Sequence or Path CountSame techniqueTree Enumeration Pattern Recognition
Everyday at school, Jo climbs a flight of 6 stairs. Jo can take the stairs 1, 2, or 3 at a time. For example, Jo could c…
- AMC 8 2018 #19 Gr 4Similar level
Same techniqueTree Enumeration Pattern RecognitionIn a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells bel…
- AMC 8 2020 #22 Gr 4Similar level
Same techniqueTree EnumerationWhen a positive integer N is fed into a machine, the output is a number calculated according to the rule shown below. Fo…