2013 AMC 10B Problem 21

A sequence starting with $u,v$ has seventh term $5u+8v$. For two sequences $(a_1,a_2)$ and $(b_1,b_2)$ with different first terms, assume $a_1\lt b_1$. Then $5a_1+8a_2=5b_1+8b_2$, so $5(b_1-a_1)=8(a_2-b_2)$. Since $5$ and $8$ are relatively prime, $b_1-a_1$ is at least $8$, and then nondecreasing order gives $b_2\ge b_1\ge a_1+8$. The smallest construction is $a_1=0$, $b_1=b_2=8$, and $a_2=13$. This gives $N=5\cdot0+8\cdot13=104$. Thus, the correct answer is C .