Competition · AMC preparation · step 4 of 4
AMC 8 · 2003 · #18
Grade 4 counting
Pick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The picture is the problem, so Tool #1 (Draw a Diagram) does the heavy lifting: label each dot with its shortest-edge distance from Sarah (1, 2, 3, or ∞ if disconnected). Tool #7 (Identify Subproblems) then splits "not invited" into two clean buckets: dots in a different piece of the graph from Sarah, and dots inside Sarah's piece but too far away (distance ≥ 3). Tool #16 (Complement) gives a fast cross-check: instead of recounting the not-invited at the end, we can also verify by counting the invited (distance 1 and 2) and subtracting from 20. Both routes must give the same answer.
Label Sarah's direct friends
Label every dot with its distance from Sarah: she connects directly to 8 dots, so those are the distance-1 friends.
Annotating the given picture with a distance label on each dot turns "who is invited?" into a routine sort.
4.OA.A.3Draw A DiagramFind the friends of friends
Step one edge further from each friend to the friends-of-friends: that distance-2 set holds 6 dots, all invited.
Step one edge further from each direct friend; any newly reached dot is a friend-of-a-friend.
4.OA.A.3Draw A DiagramSplit the uninvited into two groups
Split "not invited" into two disjoint buckets: A, dots with no path to Sarah; B, dots in her component at distance 3 or more.
Breaking a count into disjoint buckets is the standard "divide and conquer" move.
3.OA.D.8Identify SubproblemsCount the disconnected dots
Bucket A (disconnected): the lower-left triangle l-m-n is 3 dots and the lone dot a is 1, so |A| = 4.
A separate piece of the graph cannot reach Sarah no matter how many edges you walk, so every dot in it is uninvited.
3.OA.D.8Identify SubproblemsCount the too-far dots
Bucket B (too far): along the chain e-f-g-h-i-j, the middle dots g and h sit at distance 3, so |B| = 2.
Walking outward from Sarah, the chain runs out of "close" labels exactly where g and h sit.
In the long chain of dots that runs across Sarah's network, the two dots sitting in the very middle are left off the guest list, even though the dot on each side of them is invited.
▸ Why?
Sarah's guest list is built from exactly two rounds of friendship — her own friends, and then the friends of those friends — so a classmate first reached only on a third round is not on it.
▸ Why?
Each middle dot's closest tie to Sarah runs through a friend-of-a-friend, so the shortest string of friendship links joining Sarah to that dot is three links long — one longer than any invited dot's.
▸ Why?
Sorting every classmate by the length of their shortest friendship chain to Sarah drops each dot into exactly one distance level, with no dot in two levels and none skipped; the invited dots fill the one-link and two-link levels, so a three-link dot lands in the first level that is left out.
▸ Why?
A classmate who is neither one of Sarah's own friends nor a friend of one of those friends matches neither named group, so they belong to the leftover 'not invited' group instead.
▸ Why?
The whole class splits with no gaps and no overlaps into the invited part and the not-invited part, so anyone standing outside the invited part is automatically inside the other.
Add the two groups
Add the disjoint buckets: 4 + 2 = 6, choice (D).
Disjoint buckets add directly — no double-counting to worry about.
3.OA.D.8Identify SubproblemsCheck with the complement
Complement check: 8 + 6 = 14 invited, so 20 - 14 = 6 are uninvited — same answer.
Counting invited and subtracting from 20 is the complement check — it confirms the 6 found by direct counting.
4.OA.A.3Change Focus Count The ComplementLabel every dot with how many steps it is from Sarah, then sort: distance 1 and 2 get invited, everything else does not — this AMC 8 graph puzzle becomes a Grade 4 sort-and-add exercise giving 6.
- Label Sarah's direct friends
- Find the friends of friends
- Split the uninvited into two groups
- Count the disconnected dots
- Count the too-far dots
- Add the two groups
- Check with the complement
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