# 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, $0\le x\le b-1$$0\le y\le b-1$, and $0\le z\le b-1$. Thus, the sum $x+y+z$ has the property $0\le x+y+z\le b-1+b-1+b-1=3b-3$.

## Case 1, $x+y+z

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

## Case 2, $b\le x+y+z\le 3b-3$

Let ${d}_{1}=⌊\frac{x+y+z}{b}⌋$ and ${d}_{0}=x+y+z-{d}_{1}b$. Then $0\le \frac{x+y+z}{b}\le 3-\frac{3}{b}$ and $0\le ⌊\frac{x+y+z}{b}⌋\le 3-⌊\frac{3}{b}⌋$. Because b ≥ 2, $⌊\frac{3}{b}⌋=0$. It follows that $0\le ⌊\frac{x+y+z}{b}⌋\le 3$ and $0\le {d}_{1}\le 3$.

Using the property of the ceiling function:

$0\le x+y+z-\frac{x+y+z}{b}b\le x+y+z-⌊\frac{x+y+z}{b}⌋b$ $\frac{x+y+z}{b}-1<⌊\frac{x+y+z}{b}⌋$

It follows that $\frac{x+y+z}{b}b-b<⌊\frac{x+y+z}{b}⌋b$ or that $x+y+z-b<⌊\frac{x+y+z}{b}⌋b$. Now multiply −1 to both sides to get $-\left(x+y+z\right)+b>-⌊\frac{x+y+z}{b}⌋b$ and $b>x+y+z-⌊\frac{x+y+z}{b}⌋b$. Since $x+y+z$ are integers, $x+y+z-⌊\frac{x+y+z}{b}⌋b\le b-1$. Thus, $0\le x+y+z-⌊\frac{x+y+z}{b}⌋b\le b-1$ and $0\le {d}_{0}\le b-1$.

Given the above, the sum x + y + z can be expressed as $x+y+z={d}_{1}b+{d}_{0}$ where $0\le {d}_{1}\le 3$ and $0\le {d}_{0}\le b-1$. 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.