Difference between revisions of "2013 AMC 12B Problems/Problem 14"
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==Solution== | ==Solution== | ||
Let the first two terms of the first sequence be <math>x_{1}</math> and <math>x_{2}</math> and the first two of the second sequence be <math>y_{1}</math> and <math>y_{2}</math>. Computing the seventh term, we see that <math>5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}</math>. Note that this means that <math>x_{1}</math> and <math>x_{2}</math> must have the same value modulo 8. To minimize, let one of them be 0; WLOG assume that <math>x_{1} = 0</math>. Thus, the smallest possible value of <math>y_{1}</math> is <math>8</math>; since the sequences are non-decreasing <math>y_{2} \ge 8</math>. To minimize, let <math>y_{2} = 8</math>. Thus, <math>5y_{1} + 8y_{2} = 40 + 64 = \boxed{\textbf{(C) }104}</math>. | Let the first two terms of the first sequence be <math>x_{1}</math> and <math>x_{2}</math> and the first two of the second sequence be <math>y_{1}</math> and <math>y_{2}</math>. Computing the seventh term, we see that <math>5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}</math>. Note that this means that <math>x_{1}</math> and <math>x_{2}</math> must have the same value modulo 8. To minimize, let one of them be 0; WLOG assume that <math>x_{1} = 0</math>. Thus, the smallest possible value of <math>y_{1}</math> is <math>8</math>; since the sequences are non-decreasing <math>y_{2} \ge 8</math>. To minimize, let <math>y_{2} = 8</math>. Thus, <math>5y_{1} + 8y_{2} = 40 + 64 = \boxed{\textbf{(C) }104}</math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=B|num-b=13|num-a=15}} |
Revision as of 18:06, 22 February 2013
Problem
Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is . What is the smallest possible value of ?
Solution
Let the first two terms of the first sequence be and and the first two of the second sequence be and . Computing the seventh term, we see that . Note that this means that and must have the same value modulo 8. To minimize, let one of them be 0; WLOG assume that . Thus, the smallest possible value of is ; since the sequences are non-decreasing . To minimize, let . Thus, .
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 |