2013 AMC 10B Problem 21

Let the first two terms be $a$ and $b$. The sequence is
$$a,\ b,\ a+b,\ a+2b,\ 2a+3b,\ 3a+5b,\ 5a+8b,$$
so $N = 5a + 8b$. Non-decreasing requires $a \le b$ (after that, every term is a sum of nonnegative terms, so the condition holds automatically).

Suppose $(a, b)$ and $(a', b')$ with $a < a'$ both give $N$: $5a + 8b = 5a' + 8b'$, so $5(a' - a) = 8(b - b')$. Since $\gcd(5, 8) = 1$, $a' - a$ is a multiple of $8$ and $b - b'$ the same multiple of $5$. The smallest jump is $a' = a + 8$, $b' = b - 5$.

The second pair must still be non-decreasing: $a + 8 \le b - 5$, i.e. $b \ge a + 13$. To minimize $N = 5a + 8b$ take $a = 0$ and $b = 13$, giving $N = 104$, with the partner pair $(8, 8)$.

Check: $0, 13, 13, 26, 39, 65, 104$ and $8, 8, 16, 24, 40, 64, 104$. Both are non-decreasing with different first terms. Any valid configuration has $b \ge 13$, so $N \ge 8 \cdot 13 = 104$.

The answer is $\boxed{\textbf{(C)}\ 104}$.

A Fibonacci-type sequence is linear in its two starting values, so "same seventh term" is a linear Diophantine equation in $(a, b)$. Coprime coefficients force the two solutions to differ by a fixed step, and the monotonicity constraint on the shifted pair turns into a lower bound on $b$.

Test the choices from the bottom. $N = 55 = 5a + 8b$ with $a \le b$: $(a, b) = (3, 5)$ only, since $b \le 6$ and $55 - 8b$ must be a nonnegative multiple of $5$. $N = 89$: $b \in \{3, 8\}$ gives $(13, 3)$ (invalid) and $(1, 8)$; one valid pair. $N = 104$: $b \in \{3, 8, 13\}$ gives $(16, 3)$ invalid, $(8, 8)$ and $(0, 13)$; two valid pairs, so $104$ works.

• Forgetting the non-decreasing condition a <= b, which would allow smaller N such as 51 + 810 = 59 + 85 = 85.
• Miscounting the recurrence and using $N = 3a + 5b$ (the sixth term) or $8a + 13b$ (the eighth).