AMC 8 · 2020 · #16

Grade 6 algebralogic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

linear-equations-one-varsystematic-enumeration convert-to-algebraidentify-subproblems ↑ Prerequisites: linear-equations-one-varmulti-digit-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $B?$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2020 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.

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Toolkit + CCSS Solution

Understand

Restated: Six points labeled $A, B, C, D, E, F$ each stand for a different digit from $1$ to $6$. The figure has five straight lines, and on each line we add up the digits of the points it passes through. Those five line-sums add to $47$ in total. We want to know which digit $B$ stands for.

Givens: $A, B, C, D, E, F$ are the digits $1, 2, 3, 4, 5, 6$ in some order (each used exactly once); Line 1 passes through $A, B, C$ (sum $A + B + C$); Line 2 passes through $A, E, F$ (sum $A + E + F$); Line 3 passes through $C, D, E$ (sum $C + D + E$); Line 4 passes through $B, D$ (sum $B + D$); Line 5 passes through $B, F$ (sum $B + F$); All five line-sums add to $47$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: The digit assigned to point $B$

Understand

Plan

Primary tool: #15 Organize Information in More Ways

Secondary: #2 Make a Systematic List, #13 Convert to Algebra

Direct guess-and-check across $6! = 720$ digit assignments is hopeless. The big idea is to reorganize the grand total of $47$: instead of adding line-by-line, add letter-by-letter and count how many times each letter shows up across the five lines. That re-grouping (Tool #15) turns the messy sum into something almost symmetric, because every letter except $B$ appears exactly twice. A short systematic list (Tool #2) of the five lines makes the counts obvious, and a single line of algebra (Tool #13) then peels $B$ off the symmetric part.

Execute — Answer: E

#2 Make a Systematic List 4.OA.A.3 Step 1

List the five line-sums in one place.
From the figure, the digits on each of the five lines are: $A+B+C$, $A+E+F$, $C+D+E$, $B+D$, $B+F$.
Their total is $47$.

$$(A+B+C) + (A+E+F) + (C+D+E) + (B+D) + (B+F) = 47$$

💡 Writing down every line in order is a Grade 4 "set up the multi-step word problem" move — nothing fancy yet, just bookkeeping.

#15 Organize Information in More Ways 3.OA.D.9 Step 2

Reorganize the same sum letter-by-letter instead of line-by-line: count how many of the five lines each letter appears on.
$A$: lines 1, 2 ($2$ times).
$B$: lines 1, 4, 5 ($3$ times).
$C$: lines 1, 3 ($2$ times).
$D$: lines 3, 4 ($2$ times).
$E$: lines 2, 3 ($2$ times).
$F$: lines 2, 5 ($2$ times).
Every letter shows up twice — except $B$, which shows up three times.

$$2A + 3B + 2C + 2D + 2E + 2F = 47$$

💡 Re-sorting the same data by a new attribute (here: "which letter" instead of "which line") is the heart of Tool #15 — and noticing the $2,2,2,2,2,3$ pattern is a Grade 3 arithmetic-pattern observation.

#13 Convert to Algebra 6.EE.A.3 Step 3

Split the lone $3B$ into $2B + B$ so that every letter has a matching coefficient of $2$, and then factor that $2$ out of the symmetric part.
The leftover single $B$ is what makes the total odd.

$$2A + 2B + B + 2C + 2D + 2E + 2F = 2(A+B+C+D+E+F) + B = 47$$

💡 Rewriting an expression in an equivalent factored form — pulling out a common factor of $2$ — is a Grade 6 "apply properties of operations to generate equivalent expressions" skill.

#15 Organize Information in More Ways 2.NBT.B.5 Step 4

The letters $A, B, C, D, E, F$ are the digits $1$ through $6$ in some order, so $A+B+C+D+E+F = 1+2+3+4+5+6 = 21$ no matter who is who.

$$A+B+C+D+E+F = 1+2+3+4+5+6 = 21$$

💡 Reordering the unknowns into the known sequence $1,2,\dots,6$ is another use of Tool #15, and adding six small numbers is a Grade 2 "fluently add within 100" calculation.

#13 Convert to Algebra 1.OA.D.8 Step 5

Substitute $A+B+C+D+E+F = 21$ into the factored equation and solve for the unknown $B$ by subtracting from both sides.

$$2(21) + B = 47 \;\Rightarrow\; 42 + B = 47 \;\Rightarrow\; B = 47 - 42 = 5 \;\Rightarrow\; \textbf{(E)}$$

💡 Finding the missing number in $42 + \square = 47$ is a Grade 1 "unknown whole number in an addition equation" task once the hard algebra is done.

[1] #2 4.OA.A.3 List the five line-sums in one place. From the figure, the digits on each of the

[2] #15 3.OA.D.9 Reorganize the same sum letter-by-letter instead of line-by-line: count how many

[3] #13 6.EE.A.3 Split the lone $3B$ into $2B + B$ so that every letter has a matching coefficien

[4] #15 2.NBT.B.5 The letters $A, B, C, D, E, F$ are the digits $1$ through $6$ in some order, so

[5] #13 1.OA.D.8 Substitute $A+B+C+D+E+F = 21$ into the factored equation and solve for the unkno

Review

Reasonableness: $B = 5$ is one of the digits $1$--$6$, so it is a legal value, and it matches one of the answer choices. Sanity-check the total: if $B = 5$, then $2 \cdot 21 + 5 = 42 + 5 = 47$, exactly the given total. We can also see why the trick works — because every letter except $B$ appears twice, the total must be even ($2 \times 21 = 42$) plus exactly one extra copy of $B$; so $B$ is whatever pushes $42$ up to $47$, namely $5$.

Alternative: Tool #6 (Guess and Check) on the five answer choices: try each value of $B$ and check $2 \cdot 21 + B = 47$. Only $B = 5$ works (the others give $43, 44, 45, 46$, all $\ne 47$). This skips the factoring insight but still confirms (E) — a useful backup on test day.

CCSS standards used (min grade 6)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Setting up the equation that the five line-sums $(A+B+C) + (A+E+F) + (C+D+E) + (B+D) + (B+F)$ together equal $47$.)
3.OA.D.9 Identify arithmetic patterns and explain using properties of operations (Spotting the $2, 3, 2, 2, 2, 2$ count pattern — every letter appears on two lines except $B$, which appears on three.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Rewriting $2A + 3B + 2C + 2D + 2E + 2F$ as the equivalent factored form $2(A+B+C+D+E+F) + B$.)
2.NBT.B.5 Fluently add and subtract within 100 (Computing the sum of the six known digits $1 + 2 + 3 + 4 + 5 + 6 = 21$.)
1.OA.D.8 Determine the unknown whole number in an addition or subtraction equation (Solving $42 + B = 47$ for $B$ by subtracting: $B = 47 - 42 = 5$.)

⭐ This AMC 8 problem only needs Grade 6 expression-rewriting (and the trick of counting how many times each letter shows up) that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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