2020 AMC 8 Problems/Problem 16

Revision as of 13:59, 18 November 2020 by Franzliszt (talk | contribs) (Solution 1)

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 based on the points in the figure and from the information we are given. \[A+B+C\] \[A+E+F\] \[C+D+E\] \[B+D\] \[B+F\] When we add the five expressions together, and equate it to 47, we get \[2A+3B+2C+2D+2E+2F=47.\] \[2(A+B+C+D+E+F)+B=47.\] In addition, we are given that $A+B+C+D+E+F=1+2+3+4+5+6=21$, where we can assign the values for A-F randomly because we don't know their individual values. Substituting in our equation, we have \[2(A+B+C+D+E+F)+B=47.\] \[2(21)+B=47.\] \[42+B=47\] \[\boxed{\textbf{(E) }5}\] ~samrocksnature and RJ5303707

Solution 2

We can write an equation. We are given $2A+3B+2C+2D+2E+2F=47$ which simplifies to $2(A+B+C+D+E+F)+B=47$. Recall that each of $A,B,C,D,E,F$ are unique integers from $1$ to $6$. Hence, our equation simplifies to $42+B=47$ regardless of which letters equal which numbers. Now we can easily see that the answer is $\textbf{(E) }5$.

-franzlizt

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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