Difference between revisions of "2011 AMC 12A Problems/Problem 8"
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== Solution == | == Solution == | ||
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> | ||
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+ | 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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+ | 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". | ||
== 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}} |
Revision as of 20:15, 10 March 2011
Problem
In the eight term sequence ,
,
,
,
,
,
,
, the value of
is
and the sum of any three consecutive terms is
. What is
?
Solution
Let . Then from
, we find that
. From
, we then get that
. Continuing this pattern, we find
,
,
, and finally
. So
A faster technique is to assume that the problem can be solved, and thus is an invariant. Since
, assign any value to
.
is a simple value to plug in, which gives a value of
for B. The 8-term sequence is thus
. The sum of the first and the last terms is
Note that this alternate solution is not a proof. If the sum of had been asked for, this technique would have given
as an answer, when the true answer would have been "cannot be determined".
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 |