Difference between revisions of "2011 AMC 12A Problems/Problem 8"
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\textbf{(E)}\ 43 </math> | \textbf{(E)}\ 43 </math> | ||
− | == Solution == | + | ==Solution 1== |
Let <math>A=x</math>. Then from <math>A+B+C=30</math>, we find that <math>B=25-x</math>. From <math>B+C+D=30</math>, we then get that <math>D=x</math>. Continuing this pattern, we find <math>E=25-x</math>, <math>F=5</math>, <math>G=x</math>, and finally <math>H=25-x</math>. So <math>A+H=x+25-x=25 \rightarrow \boxed{\textbf{C}}</math> | Let <math>A=x</math>. Then from <math>A+B+C=30</math>, we find that <math>B=25-x</math>. From <math>B+C+D=30</math>, we then get that <math>D=x</math>. Continuing this pattern, we find <math>E=25-x</math>, <math>F=5</math>, <math>G=x</math>, and finally <math>H=25-x</math>. So <math>A+H=x+25-x=25 \rightarrow \boxed{\textbf{C}}</math> | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | Given that the sum of 3 consecutive terms is 30, we have | ||
+ | <math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math> | ||
+ | |||
+ | It follows that <math>A+B+C+D+E+F+G+H=85</math> because <math>C=5</math>. | ||
+ | |||
+ | Subtracting, we have that <math>A+H=25\rightarrow \boxed{\textbf{C}}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 3 (the tedious one)== | ||
+ | From the given information, we can deduce the following equations: | ||
+ | |||
+ | <math>A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G | ||
+ | =30</math>, and <math>F+G+H=30</math>. | ||
+ | |||
+ | We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer. | ||
+ | |||
+ | <math>(A+B)-(B+D)=25-25 \implies (A-D)=0</math> | ||
+ | |||
+ | <math>(A-D)+(D+E)=0+25 \implies (A+E)=25</math> | ||
+ | |||
+ | <math>(A+E)-(E+F+G)=25-30 \implies (A-F-G)=-5</math> (Notice how we don't use <math>D+E+F=30</math>) | ||
+ | |||
+ | <math>(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25</math> | ||
+ | |||
+ | Therefore, we have <math>A+H=25 \rightarrow \boxed{\textbf{C}}</math> | ||
+ | |||
+ | ~JinhoK | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=6tlqpAcmbz4 | ||
+ | ~Shreyas S | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=7|num-a=9|ab=A}} | {{AMC12 box|year=2011|num-b=7|num-a=9|ab=A}} | ||
+ | {{AMC10 box|year=2011|num-b=16|num-a=18|ab=A}} | ||
+ | {{MAA Notice}} |
Revision as of 14:57, 12 July 2021
Contents
Problem
In the eight term sequence , , , , , , , , the value of is and the sum of any three consecutive terms is . What is ?
Solution 1
Let . Then from , we find that . From , we then get that . Continuing this pattern, we find , , , and finally . So
Solution 2
Given that the sum of 3 consecutive terms is 30, we have and
It follows that because .
Subtracting, we have that .
Solution 3 (the tedious one)
From the given information, we can deduce the following equations:
, and .
We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer.
(Notice how we don't use )
Therefore, we have
~JinhoK
Video Solution
https://www.youtube.com/watch?v=6tlqpAcmbz4 ~Shreyas S
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 AMC 12 Problems and Solutions |
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 AMC 10 Problems and Solutions |
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