AMC 8 · 2025 · #21
Grade 6 logiccountingProblem
The Konigsberg School has assigned grades 1 through 7 to pods A through G, one grade per pod. Some of the pods are connected by walkways, as shown in the figure below. The school noticed that each pair of connected pods has been assigned grades differing by 2 or more grade levels. (For example, grades 1 and 2 will not be in pods directly connected by a walkway.) What is the sum of the grade levels assigned to pods C,E, and F?
Pick an answer.
AMC 8 2025 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: Seven pods $A$ through $G$ need to be assigned the seven grade levels $1$ through $7$, one grade per pod. Some pods are joined by walkways, and any two pods joined directly must receive grades that differ by at least $2$. Find the sum $g(C) + g(E) + g(F)$.
Givens: Seven pods $A, B, C, D, E, F, G$ each get a different grade from $\{1, 2, 3, 4, 5, 6, 7\}$; Walkway list from the figure: $A$-$B$, $A$-$C$, $A$-$F$, $A$-$G$, $B$-$C$, $B$-$F$, $C$-$D$, $C$-$E$, $C$-$F$, $D$-$E$, $E$-$F$, $F$-$G$; For every walkway pair $(X, Y)$: $|g(X) - g(Y)| \ge 2$; Answer choices: (A) $12$, (B) $13$, (C) $14$, (D) $15$, (E) $16$
Unknowns: The single value of $g(C) + g(E) + g(F)$
Understand
Restated: Seven pods $A$ through $G$ need to be assigned the seven grade levels $1$ through $7$, one grade per pod. Some pods are joined by walkways, and any two pods joined directly must receive grades that differ by at least $2$. Find the sum $g(C) + g(E) + g(F)$.
Givens: Seven pods $A, B, C, D, E, F, G$ each get a different grade from $\{1, 2, 3, 4, 5, 6, 7\}$; Walkway list from the figure: $A$-$B$, $A$-$C$, $A$-$F$, $A$-$G$, $B$-$C$, $B$-$F$, $C$-$D$, $C$-$E$, $C$-$F$, $D$-$E$, $E$-$F$, $F$-$G$; For every walkway pair $(X, Y)$: $|g(X) - g(Y)| \ge 2$; Answer choices: (A) $12$, (B) $13$, (C) $14$, (D) $15$, (E) $16$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #1 Draw a Diagram, #15 Organize Information in More Ways
The graph already gives a picture, so Tool #1 just means "read the diagram carefully and list every walkway." Tool #15 reorganizes that walkway list into a degree count per pod, which exposes that pods $C$ and $F$ are the most constrained (degree $5$). Tool #3 then drives the whole solution: for the highly constrained pods, most candidate grades are eliminated by a simple counting argument (a pod with $5$ neighbors needs $5$ usable grades among the other $6$, and only grades $1$ and $7$ leave that many), and the same elimination logic then forces $g(D)$, $g(G)$, and finally $g(E)$ one at a time.
Execute — Answer: A
K.MD.B.3 Step 1 - First, read the figure carefully and list every walkway.
- There are $12$ walkways: $A$-$B$, $A$-$C$, $A$-$F$, $A$-$G$, $B$-$C$, $B$-$F$, $C$-$D$, $C$-$E$, $C$-$F$, $D$-$E$, $E$-$F$, $F$-$G$.
💡 Reading a picture and sorting connections into a list is the same "classify and count" move kindergarteners do.
K.MD.B.3 Step 2 - Reorganize the walkway list into a degree count (how many neighbors each pod has).
- This puts the hardest pods at the top of the to-do list.
- The degree-$5$ pods $C$ and $F$ are the most constrained.
💡 Re-sorting the same information by "how many neighbors" lets us see who is most pinned down — still kindergarten counting.
6.NS.C.7 Step 3 - Eliminate impossible grades for a degree-$5$ pod.
- If pod $C$ has grade $k$, then its $5$ neighbors must use $5$ different grades from $\{1,\dots,7\}\setminus\{k-1, k, k+1\}$.
- Counting how big that allowed set is for each choice of $k$ shows only $k=1$ and $k=7$ leave $5$ usable grades.
💡 Comparing $|k - n| \ge 2$ across choices of $k$ is exactly Grade 6 absolute-value reasoning.
6.NS.C.7 Step 4 - The same argument applies to pod $F$ (also degree $5$).
- So $\{g(C), g(F)\} = \{1, 7\}$.
- By symmetry the final sum is the same in both cases; take $g(C) = 1$ and $g(F) = 7$.
💡 After eliminating everything else, only the two extreme grades survive for both highly connected pods.
6.NS.C.7 Step 5 - Place grade $6$.
- Since $g(F) = 7$, no neighbor of $F$ may have grade $6$.
- The pods adjacent to $F$ are $\{A, B, C, E, G\}$, so the only pod that can hold grade $6$ is $D$.
- Similarly, grade $2$ cannot touch $g(C) = 1$, and the only pod not adjacent to $C$ is $G$, so $g(G) = 2$.
💡 Each extreme grade ($2$ next to $1$, $6$ next to $7$) has exactly one legal home — the non-neighbor.
6.NS.C.7 Step 6 - Pin down $g(E)$.
- Pod $E$ is adjacent to $C\,(1)$, $D\,(6)$, and $F\,(7)$.
- The constraints $|g(E) - 1| \ge 2$, $|g(E) - 6| \ge 2$, $|g(E) - 7| \ge 2$ leave $g(E) \in \{3, 4\}$, and the remaining grades for $\{A, B, E\}$ are $\{3, 4, 5\}$.
- If $g(E) = 3$ then $\{g(A), g(B)\} = \{4, 5\}$, but $A$-$B$ is a walkway and $|4 - 5| = 1$ — illegal.
- So $g(E) = 4$ (one valid completion: $g(A) = 5$, $g(B) = 3$).
💡 After eliminating each candidate using the $A$-$B$ walkway as the tie-breaker, only one grade survives for $E$.
1.OA.A.2 Step 7 Add the three requested grades.
💡 Adding three small whole numbers under $20$ is Grade 1 arithmetic.
K.MD.B.3 First, read the figure carefully and list every walkway. There are $12$ walkways K.MD.B.3 Reorganize the walkway list into a degree count (how many neighbors each pod has 6.NS.C.7 Eliminate impossible grades for a degree-$5$ pod. If pod $C$ has grade $k$, then 6.NS.C.7 The same argument applies to pod $F$ (also degree $5$). So ${g(C), g(F)} = {1 6.NS.C.7 Place grade $6$. Since $g(F) = 7$, no neighbor of $F$ may have grade $6$. The po 6.NS.C.7 Pin down $g(E)$. Pod $E$ is adjacent to $C\,(1)$, $D\,(6)$, and $F\,(7)$. The co 1.OA.A.2 Add the three requested grades. Review
Reasonableness: A complete legal assignment $g(C){=}1,\ g(G){=}2,\ g(B){=}3,\ g(E){=}4,\ g(A){=}5,\ g(D){=}6,\ g(F){=}7$ can be checked against all $12$ walkways: every connected pair differs by at least $2$ (the tightest being $A$-$B$ with $|5-3|=2$ and $C$-$D$ with $|1-6|=5$, all $\ge 2$). The sum $1 + 4 + 7 = 12$ matches choice (A). The opposite case $g(C){=}7,\ g(F){=}1$ gives the symmetric assignment with the same sum $7+4+1=12$ — consistent and unique.
Alternative: Tool #6 (Guess and Check) on the multiple-choice values: only triples $(g(C), g(E), g(F))$ summing to a candidate, with $\{g(C), g(F)\}=\{1,7\}$ forced, leave $g(E)$ pinned. Sum $13$ would need $g(E)=5$, but $|5-6|=1$ violates the $E$-$D$ walkway. Sum $14$ needs $g(E)=6$, but then $g(D)$ cannot be $6$. Sums $15$ and $16$ require $g(E) \ge 7$, impossible. Only $g(E) = 4$ and total $12$ survives — choice (A).
CCSS standards used (min grade 6)
K.MD.B.3Classify objects into given categories and count the numbers in each (Reading the figure to list the $12$ walkways and then sorting them into a degree count per pod.)6.NS.C.7Understand ordering and absolute value of rational numbers (Applying the rule $|g(X) - g(Y)| \ge 2$ to eliminate impossible grades for the highly connected pods $C$ and $F$, then for $D$, $G$, and $E$.)1.OA.A.2Solve word problems involving three whole numbers whose sum is within 20 (Adding $g(C) + g(E) + g(F) = 1 + 4 + 7 = 12$ at the end.)
⭐ This AMC 8 problem only needs Grade 6 absolute-value reasoning ("two grades must differ by at least 2") you already know!
⭐ This AMC 8 problem only needs Grade 6 absolute-value reasoning ("two grades must differ by at least 2") you already know!
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