Difference between revisions of "2013 AMC 12B Problems/Problem 14"

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<math>\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273</math>
 
<math>\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273</math>
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==Solution==
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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>.

Revision as of 15:52, 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 $N$. What is the smallest possible value of $N$ ?

$\textbf{(A)}\ 55 \qquad \textbf{(B)}\ 89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273$

Solution

Let the first two terms of the first sequence be $x_{1}$ and $x_{2}$ and the first two of the second sequence be $y_{1}$ and $y_{2}$. Computing the seventh term, we see that $5x_{1} + 8x_{2} = 5y_{1} + 8y_{2}$. Note that this means that $x_{1}$ and $x_{2}$ must have the same value modulo 8. To minimize, let one of them be 0; WLOG assume that $x_{1} = 0$. Thus, the smallest possible value of $y_{1}$ is $8$; since the sequences are non-decreasing $y_{2} \ge 8$. To minimize, let $y_{2} = 8$. Thus, $5y_{1} + 8y_{2} = 40 + 64 = \boxed{\textbf{(C) }104}$.

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