Problem 16 · 2020 AMC 8
Medium
Algebra & Patterns
sum-constraintsubstitution
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Answer: E — B = 5.
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Hint 1 of 2
Don't chase individual line-sums. When you add all five sums together, each digit gets counted once for every line it sits on. So ask: how many lines pass through each point?
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Hint 2 of 2
Every point is on 2 lines except B, which is the busy crossing on 3. So the grand total counts each digit twice, plus B one extra time: 2(A+B+C+D+E+F) + B = 47.
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Approach: count incidences — each digit appears once per line it's on
- Adding all five line-sums counts each point once per line through it. Reading the figure, A, C, D, E, F lie on 2 lines each, and B (the busy crossing) lies on 3. So the grand total is 2(A+B+C+D+E+F) + B = 47.
- The digits 1–6 are used once each, so A+B+C+D+E+F = 1+2+3+4+5+6 = 21. Then 2(21) + B = 42 + B = 47.
- B = 5.
- Why this transfers: for “sum of all the line totals” problems, count incidences — a point on k lines contributes its value k times. The whole figure collapses to one equation, no casework on which digit goes where.
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