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2011 AMC 12A Problems/Problem 8: Difference between revisions

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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>
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) ==
== Solution 5 ==


Assume the sequence is <cmath>15,10,5,15,10,5,15,10</cmath>.
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>.


Thus, <math>15+10=\boxed{25}</math> or option <math>\boxed{\mathbf{(C )}25}</math>
~eevee9406


~SirAppel
== Solution 6 ==
 
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==
==Note==

Latest revision as of 17:40, 2 March 2025

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 deduce 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

Solution 4 (the cheap one)

Since all of the answer choices are constants, it shouldn't matter what we pick $A$ and $B$ to be, so let $A = 20$ and $B = 5$. Then $D = 30 - B -C = 20$, $E = 30 - D - C = 5$, $F = 30 - D - E =5$, and so on until we get $H = 5$. Thus $A + H = \boxed{\mathbf{(C)}25}$

Solution 5

Notice that the period of the sequence is $3$ as given. (If this isn't clear we can show an example: $A+B+C=B+C=D$ $\Leftrightarrow$ $A=D$). Then $A=D$ and $H=B$, so $A+H=D+B=D+B+C-C=30-5=\boxed{\textbf{(C)}\;25}$.

~eevee9406

Solution 6

Since the period of the sequence is $3$ \[A=D=G, B=E=H, C=F\] $A+B+C=30$ because any three consecutive terms sum to $30$. Since $C=5$ \[A+B=25\] Since $B=H$ \[A+B=A+H=\fbox{(C) 25}\]

~sid2012 [1]

Note

Something useful to shorten a lot of the solutions above is to notice \[5 + D + E = D + E + F\] 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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