Difference between revisions of "2020 AMC 8 Problems/Problem 16"
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===Solution 2=== | ===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>. | 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 WhyMath== | ||
| + | https://youtu.be/1ldTmo4J7Es | ||
| + | |||
| + | ~savannahsolver | ||
===Video Solution=== | ===Video Solution=== | ||
Revision as of 17:20, 26 February 2021
Each of the points 
and 
in the figure below represents a different digit from 
to 
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 
What is the digit represented by 
![[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]](http://latex.artofproblemsolving.com/d/8/7/d878779f0718f46b33d37199ff7b7df96df2948e.png)
Contents
Solutions
Solution 1
We can form the following expressions for the sum along each line:
![\[\begin{dcases}A+B+C\\A+E+F\\C+D+E\\B+D\\B+F\end{dcases}\]](http://latex.artofproblemsolving.com/c/e/d/ced75e75ee703776845d4ca9666de8729ada010b.png)
Adding these together, we must have 
, i.e. 
. Since 
are unique integers between and 
, we obtain 
(where the order doesn't matter as addition is commutative), so our equation simplifies to 
. This means 
.
~RJ5303707
Solution 2
Following the first few steps of Solution 1, we have 
. Because an even number (
) subtracted from an odd number (47) is always odd, we know that 
is odd, showing that 
is odd. Now we know that is either 1, 3, or 5. If we try 
, we get 
which is not true. Testing 
, we get 
, which is also not true. Therefore, we have .
Video Solution by WhyMath
~savannahsolver
Video Solution
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. 
