Difference between revisions of "2020 AMC 8 Problems/Problem 16"
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Each of the points <math>A,B,C,D,E,</math> and <math>F</math> in the figure below represents a different digit from <math>1</math> to <math>6.</math> 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 <math>47.</math> What is the digit represented by <math>B?</math> | Each of the points <math>A,B,C,D,E,</math> and <math>F</math> in the figure below represents a different digit from <math>1</math> to <math>6.</math> 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 <math>47.</math> What is the digit represented by <math>B?</math> | ||
<asy> | <asy> | ||
Line 46: | Line 45: | ||
<math>\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5</math> | <math>\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5</math> | ||
− | ==Solution== | + | ==Solutions== |
+ | |||
+ | ===Solution 1=== | ||
We can form the following expressions for the sum along each line: | We can form the following expressions for the sum along each line: | ||
<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 | ||
+ | |||
+ | ===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>. | ||
− | ==Video Solution== | + | ==Video Solution by WhyMath== |
https://youtu.be/1ldTmo4J7Es | https://youtu.be/1ldTmo4J7Es | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ===Video Solution=== | ||
+ | https://youtu.be/VnOecUiP-SA | ||
==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}} |
Latest revision as of 18:20, 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
Contents
Solutions
Solution 1
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
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.