Difference between revisions of "2020 AMC 8 Problems/Problem 16"
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==Solution 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> | + | 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> <cmath>\boxed{\textbf{(E) }5}</cmath> |
~samrocksnature | ~samrocksnature |
Revision as of 00:35, 18 November 2020
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 1
We notice that 3 lines pass through B, and 2 lines pass through all other points. In addition, we are given that . This means that
~samrocksnature
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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