Difference between revisions of "2011 AMC 12A Problems/Problem 8"

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\textbf{(E)}\ 43 </math>
 
\textbf{(E)}\ 43 </math>
  
== Solution ==
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==Solution 1==
===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===
 
  
A faster technique is to assume that the problem can be solved, and thus <math>A+H</math> is an invariant.  Since <math>A + B + 5 = 30</math>, assign any value to <math>A</math>.  <math>10</math> is a simple value to plug in, which gives a value of <math>15</math> for B.  The 8-term sequence is thus <math>10, 15, 5, 10, 15, 5, 10, 15</math>.  The sum of the first and the last terms is <math>25\rightarrow \boxed{\textbf{C}}</math>
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==Solution 2==
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Given that the sum of 3 consecutive terms is 30, we have
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<math>(A+B+C)+(C+D+E)+(F+G+H)=90</math> and <math>(B+C+D)+(E+F+G)=60</math>
  
Note that this alternate solution is not a proof.  If the sum of <math>A+G</math> had been asked for, this technique would have given <math>20</math> as an answer, when the true answer would have been "cannot be determined".
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It follows that <math>A+B+C+D+E+F+G+H=85</math> because <math>C=5</math>.
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Subtracting, we have that <math>A+H=25\rightarrow \boxed{\textbf{C}}</math>.
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 +
 
 +
==Solution 3 (the tedious one)==
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From the given information, we can deduct the following equations:
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<math>A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G
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=30</math>, and <math>F+G+H=30</math>.
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 +
We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer.
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<math>(A+B)-(B+D)=25-25 \implies (A-D)=0</math>
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<math>(A-D)+(D+E)=0+25 \implies (A+E)=25</math>
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<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>)
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<math>(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25</math>
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Therefore, we have <math>A+H=25 \rightarrow \boxed{\textbf{C}}</math>
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 +
~JinhoK
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==Video Solution==
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https://www.youtube.com/watch?v=6tlqpAcmbz4
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~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}}
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{{MAA Notice}}

Revision as of 14:53, 5 May 2021

Problem

In the eight term sequence $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, the value of $C$ is $5$ and the sum of any three consecutive terms is $30$. What is $A+H$?

$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$

Solution 1

Let $A=x$. Then from $A+B+C=30$, we find that $B=25-x$. From $B+C+D=30$, we then get that $D=x$. Continuing this pattern, we find $E=25-x$, $F=5$, $G=x$, and finally $H=25-x$. So $A+H=x+25-x=25 \rightarrow \boxed{\textbf{C}}$


Solution 2

Given that the sum of 3 consecutive terms is 30, we have $(A+B+C)+(C+D+E)+(F+G+H)=90$ and $(B+C+D)+(E+F+G)=60$

It follows that $A+B+C+D+E+F+G+H=85$ because $C=5$.

Subtracting, we have that $A+H=25\rightarrow \boxed{\textbf{C}}$.


Solution 3 (the tedious one)

From the given information, we can deduct the following equations:

$A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G =30$, and $F+G+H=30$.

We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer.

$(A+B)-(B+D)=25-25 \implies (A-D)=0$

$(A-D)+(D+E)=0+25 \implies (A+E)=25$

$(A+E)-(E+F+G)=25-30 \implies (A-F-G)=-5$ (Notice how we don't use $D+E+F=30$)

$(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25$

Therefore, we have $A+H=25 \rightarrow \boxed{\textbf{C}}$

~JinhoK

Video Solution

https://www.youtube.com/watch?v=6tlqpAcmbz4 ~Shreyas S

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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