Difference between revisions of "2011 AMC 12A Problems/Problem 8"
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\textbf{(E)}\ 43 </math> | \textbf{(E)}\ 43 </math> | ||
− | + | ==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 | 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> | <math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math> | ||
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− | + | ==Solution 3 (the tedious one)== | |
− | From the given information, we can | + | 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 | <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>. | =30</math>, and <math>F+G+H=30</math>. | ||
− | We can then | + | 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+B)-(B+D)=25-25 \implies (A-D)=0</math> | ||
− | <math>(A-D)+(D+E)=0+25 \implies (A+E)=25</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> | + | <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> | <math>(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25</math> | ||
− | Therefore, we have | + | Therefore, we have <math>A+H=25 \rightarrow \boxed{\textbf{C}}</math> |
~JinhoK | ~JinhoK | ||
+ | |||
+ | == Solution 4 (the cheap one) == | ||
+ | |||
+ | Since all of the answer choices are constants, it shouldn't matter what we pick <math>A</math> and <math>B</math> to be, so let <math>A = 20</math> and <math>B = 5</math>. Then <math>D = 30 - B -C = 20</math>, <math>E = 30 - D - C = 5</math>, <math>F = 30 - D - E =5</math>, and so on until we get <math>H = 5</math>. Thus <math>A + H = \boxed{\mathbf{(C)}25}</math> | ||
+ | |||
+ | == Solution 5 (assumption) == | ||
+ | |||
+ | Assume the sequence is <cmath>15,10,5,15,10,5,15,10</cmath>. | ||
+ | |||
+ | Thus, <math>15+10=\boxed{25}</math> or option <math>\boxed{\mathbf{(C )}25}</math> | ||
+ | |||
+ | ~SirAppel | ||
+ | |||
+ | == Solution 6 == | ||
+ | |||
+ | Notice that the period of the sequence is <math>3</math> as given. (If this isn't clear we can show an example: <math>A+B+C=B+C=D</math> <math>\Leftrightarrow</math> <math>A=D</math>). Then <math>A=D</math> and <math>H=B</math>, so <math>A+H=D+B=D+B+C-C=30-5=\boxed{\textbf{(C)}\;25}</math>. | ||
+ | |||
+ | ~eevee9406 | ||
+ | |||
+ | == Solution 7 == | ||
+ | |||
+ | Since the period of the sequence is <math>3</math> <cmath>A=D=G, B=E=H, C=F</cmath> | ||
+ | <math>A+B+C=30</math> because any three consecutive terms sum to <math>30</math>. | ||
+ | Since <math>C=5</math> <cmath>A+B=25</cmath> | ||
+ | Since <math>B=H</math> <cmath>A+B=A+H=\fbox{(C) 25}</cmath> | ||
+ | |||
+ | ~sid2012 [https://artofproblemsolving.com/wiki/index.php/User:Sid2012] | ||
+ | |||
+ | ==Note== | ||
+ | Something useful to shorten a lot of the solutions above is to notice <cmath>5 + D + E = D + E + F</cmath> so F = 5 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=6tlqpAcmbz4 | ||
+ | ~Shreyas S | ||
+ | |||
+ | ==Podcast Solution== | ||
+ | |||
+ | https://www.buzzsprout.com/56982/episodes/383730 (Episode starts with a solution to this question) | ||
+ | —[[User:wescarroll|wescarroll]] | ||
== 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}} | {{MAA Notice}} |
Latest revision as of 13:05, 29 September 2024
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
Solution 4 (the cheap one)
Since all of the answer choices are constants, it shouldn't matter what we pick and to be, so let and . Then , , , and so on until we get . Thus
Solution 5 (assumption)
Assume the sequence is .
Thus, or option
~SirAppel
Solution 6
Notice that the period of the sequence is as given. (If this isn't clear we can show an example: ). Then and , so .
~eevee9406
Solution 7
Since the period of the sequence is because any three consecutive terms sum to . Since Since
~sid2012 [1]
Note
Something useful to shorten a lot of the solutions above is to notice so F = 5
Video Solution
https://www.youtube.com/watch?v=6tlqpAcmbz4 ~Shreyas S
Podcast Solution
https://www.buzzsprout.com/56982/episodes/383730 (Episode starts with a solution to this question) —wescarroll
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.