2013 AMC 10B Problems/Problem 21

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 nondecreasing $y_{2} \ge 8$ . To minimize, let $y_{2} = 8$ . Thus, $5y_{1} + 8y_{2} = 40 + 64 = \boxed{\textbf{(C) }104}$ .

I don't know why the person got the answer above, but the answer seems to be 55. if $x_{1}$ and $x_{2}$ are 11 and 0, respectively, and $y_{1}$ and $y_{2}$ are 3 and 5, respectively, the answer would be \boxed{\textbf{(A) }55}$.

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2013 AMC 10B Problems/Problem 21

Problem

Solution