Difference between revisions of "2020 AMC 8 Problems/Problem 16"
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<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 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> | ||
Line 54: | Line 54: | ||
~RJ5303707 | ~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>. | 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 by North America Math Contest Go Go Go== | + | ==Video Solution by NiuniuMaths (Easy to understand!)== |
+ | https://www.youtube.com/watch?v=bHNrBwwUCMI | ||
+ | |||
+ | ~NiuniuMaths | ||
+ | |||
+ | == Video Solution by Math-X (First understand the problem!!!) == | ||
+ | https://youtu.be/UnVo6jZ3Wnk?si=n-vOSvxYPmuhVmaq&t=2574 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | == Video Solution (CLEVER MANIPULATIONS!!!) == | ||
+ | https://youtu.be/W8pib6O_6xA | ||
+ | |||
+ | ~<i>Education, the Study of Everything</i> | ||
+ | |||
+ | == Video Solution by North America Math Contest Go Go Go == | ||
https://www.youtube.com/watch?v=hwCb64F34XE | https://www.youtube.com/watch?v=hwCb64F34XE | ||
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~North America Math Contest Go Go Go | ~North America Math Contest Go Go Go | ||
− | ==Video Solution by WhyMath== | + | == Video Solution by WhyMath == |
https://youtu.be/1ldTmo4J7Es | https://youtu.be/1ldTmo4J7Es | ||
~savannahsolver | ~savannahsolver | ||
− | ==Video Solution== | + | == Video Solution == |
https://www.youtube.com/watch?v=a3Z7zEc7AXQ | https://www.youtube.com/watch?v=a3Z7zEc7AXQ | ||
-LOUISGENIUS | -LOUISGENIUS | ||
− | ==Video Solution by Interstigation== | + | == Video Solution by Interstigation == |
https://youtu.be/YnwkBZTv5Fw?t=728 | https://youtu.be/YnwkBZTv5Fw?t=728 | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
− | + | == 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 16:35, 26 January 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution by NiuniuMaths (Easy to understand!)
- 5 Video Solution by Math-X (First understand the problem!!!)
- 6 Video Solution (CLEVER MANIPULATIONS!!!)
- 7 Video Solution by North America Math Contest Go Go Go
- 8 Video Solution by WhyMath
- 9 Video Solution
- 10 Video Solution by Interstigation
- 11 Video Solution by OmegaLearn
- 12 See also
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 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 NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=bHNrBwwUCMI
~NiuniuMaths
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=n-vOSvxYPmuhVmaq&t=2574
~Math-X
Video Solution (CLEVER MANIPULATIONS!!!)
~Education, the Study of Everything
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=hwCb64F34XE
~North America Math Contest Go Go Go
Video Solution by WhyMath
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=a3Z7zEc7AXQ -LOUISGENIUS
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=728
~Interstigation
Video Solution by OmegaLearn
https://youtu.be/sZfOjGtEtEY?t=604
~ pi_is_3.14
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.