Proof: The sum of any three single-digit numbers is at most two digits long

Here comes another trivial proof (the base must be at least 2 for the obvious reason):

Suppose x, y, z are single-digit numbers in base b ≥ 2. By the definition, 0xb10yb1, and 0zb1. Thus, the sum x+y+z has the property 0x+y+zb1+b1+b1=3b3.

Case 1, x+y+z<b

In this case, the sum can be expressed in single digit.

Case 2, bx+y+z3b3

Let d1=x+y+zb and d0=x+y+zd1b. Then 0x+y+zb33b and 0x+y+zb33b. Because b ≥ 2, 3b=0. It follows that 0x+y+zb3 and 0d13.

Using the property of the ceiling function:

0x+y+zx+y+zbbx+y+zx+y+zbb x+y+zb1<x+y+zb

It follows that x+y+zbbb<x+y+zbb or that x+y+zb<x+y+zbb. Now multiply −1 to both sides to get (x+y+z)+b>x+y+zbb and b>x+y+zx+y+zbb. Since x+y+z are integers, x+y+zx+y+zbbb1. Thus, 0x+y+zx+y+zbbb1 and 0d0b1.

Given the above, the sum x + y + z can be expressed as x+y+z=d1b+d0 where 0d13 and 0d0b1. Hence x+y+z can be expressed in two digits.

Thus in both cases, the sum of three numbers x, y, z can be at most two digits long. Q.E.D.