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

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<math>\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5</math>
 
<math>\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5</math>
  
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==Solution 1==
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==Solutions==
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We can form the following expressions based on the points in the figure and from the information we are given.
 
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<cmath>A+B+C</cmath>
 
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<cmath>A+E+F</cmath>
 
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<cmath>C+D+E</cmath>
 
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<cmath>B+D</cmath>
 
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<cmath>B+F</cmath>
 
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When we add the five expressions together, and equate it to 47, we get
 
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<cmath>2A+3B+2C+2D+2E+2F=47.</cmath>
 
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<cmath>2(A+B+C+D+E+F)+B=47.</cmath>
 
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In addition, we are given that <math>A+B+C+D+E+F=1+2+3+4+5+6=21</math>, where we can assign the values for A-F randomly because we don't know their individual values. Substituting in our equation, we have
 
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<cmath>2(A+B+C+D+E+F)+B=47.</cmath>
 
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<cmath>2(21)+B=47.</cmath>
 
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<cmath>42+B=47</cmath>
 
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<cmath>\boxed{\textbf{(E) }5}</cmath>
 
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~samrocksnature and RJ5303707
 
  
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==Solution 2==
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===Solution 1===
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We can form the following expressions for the sum along each line:
 +
<cmath>\begin{dcases}A+B+C\\A+E+F\\C+D+E\\B+D\\B+F\end{dcases}</cmath>
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Adding these together, we must have <math>2A+3B+2C+2D+2E+2F=47</math>, i.e. <math>2(A+B+C+D+E+F)+B=47</math>. Since <math>A,B,C,D,E,F</math> are unique integers between <math>1</math> and <math>6</math>, we obtain <math>A+B+C+D+E+F=1+2+3+4+5+6=21</math> (where the order doesn't matter as addition is commutative), so our equation simplifies to <math>42 + B = 47</math>. This means <math>B = \boxed{\textbf{(E) }5}</math>.
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~RJ5303707
  
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We can write an equation. We are given <math>2A+3B+2C+2D+2E+2F=47</math> which simplifies to <math>2(A+B+C+D+E+F)+B=47</math>. Recall that each of <math>A,B,C,D,E,F</math> are unique integers from <math>1</math> to <math>6</math>. Hence, our equation simplifies to <math>42+B=47</math> regardless of which letters equal which numbers. Now we can easily see that the answer is <math>\textbf{(E) }5</math>.
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===Solution 2===
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Following the first few steps of Solution 1, we have <math>2(A+C+D+E+F)+3B=47</math>. Because an even number (<math>2(A+C+D+E+F)</math>) subtracted from an odd number (47) is always odd, we know that <math>3B</math> is odd, showing that <math>B</math> is odd. Now we know that <math>B</math> is either 1, 3, or 5. If we try <math>B=1</math>, we get <math>43=47</math> which is not true. Testing <math>B=3</math>, we get <math>45=47</math>, which is also not true. Therefore, we have <math>B = \boxed{\textbf{(E) }5}</math>.
  
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-franzliszt
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===Video Solution===
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https://youtu.be/VnOecUiP-SA
  
 
==See also==
 
==See also==
 
{{AMC8 box|year=2020|num-b=15|num-a=17}}
 
{{AMC8 box|year=2020|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:41, 16 January 2021

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]

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

Solutions

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 $B = \boxed{\textbf{(E) }5}$. ~RJ5303707

Solution 2

Following the first few steps of Solution 1, we have $2(A+C+D+E+F)+3B=47$. Because an even number ($2(A+C+D+E+F)$) subtracted from an odd number (47) is always odd, we know that $3B$ is odd, showing that $B$ is odd. Now we know that $B$ is either 1, 3, or 5. If we try $B=1$, we get $43=47$ which is not true. Testing $B=3$, we get $45=47$, which is also not true. Therefore, we have $B = \boxed{\textbf{(E) }5}$.

Video Solution

https://youtu.be/VnOecUiP-SA

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 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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