Difference between revisions of "2020 AMC 8 Problems/Problem 16"
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We can write an equation. We are given <math>2A+3B+2C+2D+2E+2F=47</math> which simplifies to <math>2(A+B+C+D+E+F)+B=47</math>. Recall that each of <math>A,B,C,D,E,F</math> are unique integers from <math>1</math> to <math>6</math>. Hence, our equation simplifies to <math>42+B=47</math> regardless of which letters equal which numbers. Now we can easily see that the answer is <math>\textbf{(E) }5</math>. | We can write an equation. We are given <math>2A+3B+2C+2D+2E+2F=47</math> which simplifies to <math>2(A+B+C+D+E+F)+B=47</math>. Recall that each of <math>A,B,C,D,E,F</math> are unique integers from <math>1</math> to <math>6</math>. Hence, our equation simplifies to <math>42+B=47</math> regardless of which letters equal which numbers. Now we can easily see that the answer is <math>\textbf{(E) }5</math>. | ||
− | - | + | -franzliszt |
==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 12:03, 19 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 can form the following expressions based on the points in the figure and from the information we are given.
When we add the five expressions together, and equate it to 47, we get
In addition, we are given that
, where we can assign the values for A-F randomly because we don't know their individual values. Substituting in our equation, we have
~samrocksnature and RJ5303707
Solution 2
We can write an equation. We are given which simplifies to
. Recall that each of
are unique integers from
to
. Hence, our equation simplifies to
regardless of which letters equal which numbers. Now we can easily see that the answer is
.
-franzliszt
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.