Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "Solution 1: We notice that 3 lines pass through B, and 2 lines pass through all other points. In addition, we are given that <math>A+B+C+D+E+F=1+2+3+4+5+6=21</math>. This mean...")
 
m (→‎Problem)
 
(51 intermediate revisions by 25 users not shown)
Line 1: Line 1:
−
Solution 1: We notice that 3 lines pass through B, and 2 lines pass through all other points. In addition, we are given that <math>A+B+C+D+E+F=1+2+3+4+5+6=21</math>. This means that <cmath>2A+3B+2C+2D+2E+2F=47</cmath> <cmath>2(A+B+C+D+E+F)+B=47</cmath> <cmath>2(21)+B=47</cmath> <cmath>B=5</cmath>
+
==Problem 16==
  
−
~samrocksnature
+
Each of the points <math>A,B,C,D,E,</math> and <math>F</math> in the figure below represents a different digit from <math>1</math> to <math>6.</math> 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 <math>47.</math> What is the digit represented by <math>B?</math>
 +
 
 +
<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>
 +
 
 +
<math>\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5</math>
 +
 
 +
== Solution 1 ==
 +
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>
 +
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>.
 +
~RJ5303707
 +
 
 +
== Solution 2 ==
 +
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>.
 +
 
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=bHNrBwwUCMI
 +
 
 +
~NiuniuMaths
 +
 
 +
== Video Solution by Math-X (First understand the problem!!!) ==
 +
https://youtu.be/UnVo6jZ3Wnk?si=n-vOSvxYPmuhVmaq&t=2574
 +
 
 +
~Math-X
 +
 
 +
== Video Solution (CLEVER MANIPULATIONS!!!) ==
 +
https://youtu.be/W8pib6O_6xA
 +
 
 +
~<i>Education, the Study of Everything</i>
 +
 
 +
== Video Solution by North America Math Contest Go Go Go ==
 +
 
 +
https://www.youtube.com/watch?v=hwCb64F34XE
 +
 
 +
~North America Math Contest Go Go Go
 +
 
 +
== Video Solution by WhyMath ==
 +
https://youtu.be/1ldTmo4J7Es
 +
 
 +
~savannahsolver
 +
 
 +
== Video Solution ==
 +
https://www.youtube.com/watch?v=a3Z7zEc7AXQ
 +
-LOUISGENIUS
 +
 
 +
== Video Solution by Interstigation ==
 +
https://youtu.be/YnwkBZTv5Fw?t=728
 +
 
 +
~Interstigation
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/sZfOjGtEtEY?t=604
 +
 
 +
~ pi_is_3.14
 +
 
 +
== See also ==
 +
{{AMC8 box|year=2020|num-b=15|num-a=17}}
 +
{{MAA Notice}}

Latest revision as of 20:57, 3 September 2024

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 $B = \boxed{\textbf{(E) }5}$. ~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 by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=n-vOSvxYPmuhVmaq&t=2574

~Math-X

Video Solution (CLEVER MANIPULATIONS!!!)

https://youtu.be/W8pib6O_6xA

~Education, the Study of Everything

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=hwCb64F34XE

~North America Math Contest Go Go Go

Video Solution by WhyMath

https://youtu.be/1ldTmo4J7Es

~savannahsolver

Video Solution

https://www.youtube.com/watch?v=a3Z7zEc7AXQ -LOUISGENIUS

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=728

~Interstigation

Video Solution by OmegaLearn

https://youtu.be/sZfOjGtEtEY?t=604

~ pi_is_3.14

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_16&oldid=226410"