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

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 1

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 $B = \boxed{\textbf{(E) }5}$.

## Solution 2 (similar to Solution 1)

We can see that $A+B+C+C+D+E+E+F+A+B+D+B+F=47$. Combining like terms, we get $2A+3B+2C+2D+2E+2F=47$. Since $A, B, C, D, E,$ and $F$ are all distinct integers ranging from 1 to 6, they must sum up to $21$, so $B=47-2(21)=\boxed{\textbf{(E) }5}$.

## Video Solution

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 