2013 AMC 10B Problem 21

1. Call the first two terms $a$ and $b$ and write out the sequence:
$$a,\quad b,\quad a+b,\quad a+2b,\quad 2a+3b,\quad 3a+5b,\quad 5a+8b.$$
So the seventh term is $N = 5a + 8b$.
2. "Non-decreasing" means no term is smaller than the one before. The only step that can fail is from $a$ to $b$, so we need $b \ge a$. After that, each new term adds a number that is $0$ or more, so it never goes down.
3. Now take two sequences, starting with $(a, b)$ and $(a', b')$, where $a < a'$. They have the same seventh term:
$$5a + 8b = 5a' + 8b' \quad\Longrightarrow\quad 5(a' - a) = 8(b - b').$$
4. The right side is a multiple of $8$, and $5$ shares no factor with $8$, so $a' - a$ must be a multiple of $8$. The smallest possibility is $a' = a + 8$, and then $b' = b - 5$. (Raising $a$ by $8$ adds $40$, and lowering $b$ by $5$ takes $40$ away.)
5. The second sequence must also be non-decreasing, so $b' \ge a'$:
$$b - 5 \ge a + 8, \quad\text{so}\quad b \ge a + 13 \ge 13.$$
A bigger jump in $a$ (like $16$) only makes this worse.

$N$ cannot be below $104$: from step 5, $b \ge 13$, so $N = 5a + 8b \ge 8 \times 13 = 104$.

$N = 104$ works: take $a = 0$, $b = 13$ and its partner $a' = 8$, $b' = 8$:
$$0, 13, 13, 26, 39, 65, 104 \qquad\text{and}\qquad 8, 8, 16, 24, 40, 64, 104.$$
Both are non-decreasing, their first terms differ, and both reach $104$.

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

A sequence where each term is the sum of the two before it is completely decided by its first two terms, and every later term is a simple combination of them. "Same seventh term" becomes one equation, and because $5$ and $8$ share no factor, two solutions must differ by a fixed jump. The non-decreasing rule on the shifted sequence then gives a lower limit.

Test the answer choices from the smallest, listing pairs with $5a + 8b = N$ and $a \le b$.

• $N = 55$: $b = 5$ gives $a = 3$; $b = 0$ gives $a = 11 > b$. Only one valid pair, so $55$ fails.
• $N = 89$: $b = 3$ gives $a = 13 > b$; $b = 8$ gives $a = 5$. Only one valid pair, so $89$ fails.
• $N = 104$: $b = 3$ gives $a = 16 > b$; $b = 8$ gives $a = 8$; $b = 13$ gives $a = 0$. Two valid pairs, so $104$ works.