Difference between revisions of "2020 AMC 8 Problems/Problem 16"
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<cmath>B+F</cmath> | <cmath>B+F</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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/1ldTmo4J7Es | ||
+ | |||
+ | ~savannahsolver | ||
==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 17:39, 20 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
.
Video Solution
~savannahsolver
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.