Difference between revisions of "2020 AMC 8 Problems/Problem 16"
(→Solution) |
(→Solution) |
||
Line 50: | Line 50: | ||
<cmath>\begin{dcases}A+B+C\\A+E+F\\C+D+E\\B+D\\B+F\end{dcases}</cmath> | <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>. | 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 | ~RJ5303707 | ||
Revision as of 20:14, 24 November 2020
Contents
Problem
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
Solution
We can form the following expressions for the sum along each line: 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
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.