Suppose that someone starts with a chain letter. Each person who receives the…

Concept. In a full m-ary tree every internal vertex has exactly m children, and every vertex except the root is the child of exactly one internal vertex. Writing i for the number of internal vertices, n for the total number of vertices and l for the number of leaves, this gives the two counting identities n = m·i + 1 and l = n − i = (m − 1)·i + 1. Knowing any one of i, n or l therefore fixes the other two.

Application. Model the chain letter as a rooted tree:

• The first person is the root, and an edge is drawn from x to y whenever x sends the letter to y.

• No one receives more than one letter, so every person other than the root has exactly one parent — the structure really is a tree, with no cycles and no repeats.

• A person who forwards the letter sends it to exactly 4 people, so every internal vertex has exactly 4 children: the tree is a full m-ary tree with m = 4.

• The 100 people who read the letter but did not send it out are exactly the vertices with no children, so the number of leaves is l = 100.

Now apply the leaf identity and then the vertex identity:

1. Substitute m = 4 and l = 100 into l = (m − 1)·i + 1, which gives 100 = 3i + 1.

2. Solve for i: 3i = 100 − 1 = 99, so i = 33. These 33 internal vertices are the people who sent the letter out.

3. Substitute i = 33 into n = m·i + 1: n = 4 × 33 + 1 = 133. These 133 vertices are all the people who saw the letter, the first person included.

Cross-check by counting letters instead of vertices. The 33 senders each posted 4 letters, so 33 × 4 = 132 letters were posted. Every person except the originator received exactly one letter, so the readership is 132 + 1 = 133, and the number who read without forwarding is 133 − 33 = 100 — exactly the figure the question fixes.

Contrast. Every offered pair already differs by 100, so the 100-leaf condition alone does not separate them; the identity n = 4i + 1 does:

Result: 133 people saw the letter, and 33 of them sent it out.