Difference between revisions of "2011 AMC 12A Problems/Problem 8"
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== Solution == | == 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=== | ||
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> | 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> |
Revision as of 12:14, 30 May 2011
Contents
[hide]Problem
In the eight term sequence , , , , , , , , the value of is and the sum of any three consecutive terms is . What is ?
Solution
Solution 1
Let . Then from , we find that . From , we then get that . Continuing this pattern, we find , , , and finally . So
Solution 2
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 |