# Difference between revisions of "2020 AMC 8 Problems/Problem 16"

## 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 $B?$ $[asy] size(200); dotfactor = 10; pair p1 = (-28,0); pair p2 = (-111,213); draw(p1--p2,linewidth(1)); pair p3 = (-160,0); pair p4 = (-244,213); draw(p3--p4,linewidth(1)); pair p5 = (-316,0); pair p6 = (-67,213); draw(p5--p6,linewidth(1)); pair p7 = (0, 68); pair p8 = (-350,10); draw(p7--p8,linewidth(1)); pair p9 = (0, 150); pair p10 = (-350, 62); draw(p9--p10,linewidth(1)); pair A = intersectionpoint(p1--p2, p5--p6); dot("A", A, 2*W); pair B = intersectionpoint(p5--p6, p3--p4); dot("B", B, 2*WNW); pair C = intersectionpoint(p7--p8, p5--p6); dot("C", C, 1.5*NW); pair D = intersectionpoint(p3--p4, p7--p8); dot("D", D, 2*NNE); pair EE = intersectionpoint(p1--p2, p7--p8); dot("E", EE, 2*NNE); pair F = intersectionpoint(p1--p2, p9--p10); dot("F", F, 2*NNE); [/asy]$

$\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5$

## Solution

We can form the following expressions for the sum along each line: $$\begin{dcases}A+B+C\\A+E+F\\C+D+E\\B+D\\B+F\end{dcases}$$ Adding these together, we must have $2A+3B+2C+2D+2E+2F=47$, i.e. $2(A+B+C+D+E+F)+B=47$. Since $A,B,C,D,E,F$ are unique integers between $1$ and $6$, we obtain $A+B+C+D+E+F=1+2+3+4+5+6=21$ (where the order doesn't matter as addition is commutative), so our equation simplifies to $42 + B = 47$. This means $B = \boxed{\textbf{(E) }5}$.

~RJ5303707

## See also

 2020 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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