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

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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Organize Information in More Ways

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.

1STEP 1

List the five line-sums — A+B+C, A+E+F, C+D+E, B+D, B+F — whose total is 47.

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

2STEP 2

Recount letter by letter: each letter lands on two lines except B, which appears three times, giving 2A+3B+2C+2D+2E+2F=47.

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

3STEP 3

Split 3B into 2B+B so every letter has coefficient 2, then factor it out: 2(A+B+C+D+E+F)+B=47.

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

4STEP 4

Since A, B, C, D, E, F are the digits 1 to 6 in some order, A+B+C+D+E+F = 1+2+3+4+5+6 = 21.

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

5STEP 5

Substitute 21: 2(21)+B=47, so 42+B=47 and B=5.

2(21) + B = 47 → 42 + B = 47 → B = 47 - 42 = 5 → (E)

Answer

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 · 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 × 21 = 42) plus exactly one extra copy of B; so B is whatever pushes 42 up to 47, namely 5.

💡Key takeaway

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!

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