2026 AMC 10A Problem 15

Going down, the tree doubles in size at every level, so searching downward for $\frac{20}{27}$ could take a very long time. Going up is easy, because every node has exactly one parent. So work backwards from $\frac{20}{27}$ to the top.

1. Look at one parent and its children. In the figure, $\frac{2}{5}$ has children $\frac{2}{7}$ and $\frac{5}{7}$. Notice two things:
- both children have the same bottom number, $7 = 2 + 5$, which is the parent's top plus bottom;
- the two children's top numbers, $2$ and $5$, are exactly the parent's top and bottom.

This is true for every node, because $\frac{a}{b}$ has children $\frac{a}{a+b}$ and $\frac{b}{a+b}$.

1. Every node is less than $1$. The top node $\frac12$ is less than $1$, and each child has a top number smaller than its bottom number $a + b$. So every parent is written as smaller number over larger number.

1. How to go up one level. Start from a node $\frac{p}{q}$. Its partner (the other child of the same parent) is $\frac{q - p}{q}$. The parent is built from the two top numbers $p$ and $q - p$, with the smaller one on top. Test it on $\frac{2}{7}$: the two numbers are $2$ and $5$, so the parent is $\frac{2}{5}$, matching the figure.

1. Climb from $\frac{20}{27}$.

1. Count the levels. Each step moves up exactly one level, and after $8$ steps we reach $\frac12$ at level $0$. So $\frac{20}{27}$ is at level $8$. Check with the figure: $\frac15$ is drawn at level $3$, and $\frac{20}{27}$ is $5$ steps below it, so its level is $3 + 5 = 8$.

6. Choice (E) is out. Reading the table from the bottom up gives an actual path down the tree:
$$\tfrac12 \to \tfrac13 \to \tfrac14 \to \tfrac15 \to \tfrac16 \to \tfrac67 \to \tfrac{7}{13} \to \tfrac{7}{20} \to \tfrac{20}{27}.$$
For instance, $\frac{7}{20}$ sprouts $\frac{7}{27}$ and $\frac{20}{27}$. So the node does appear.

The answer is $\boxed{\textbf{(B)}\ 8}$.

When a process branches out going forward but has only one way back, work backwards: the path from the start to any node is forced, so climbing up finds it with no guessing. Each step up replaces the pair of numbers $(p, q)$ by $(p, q - p)$, which is the subtraction form of the Euclidean algorithm, so the numbers keep shrinking and the climb always ends at $\frac12$. In fact, every fraction between $0$ and $1$ in lowest terms appears in this tree exactly once.