# 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, $$x+y+z<b$$

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

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

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

Using the property of the ceiling function:

$$0\le x+y+z-\frac{x+y+z}{b}b\le x+y+z-\lfloor \frac{x+y+z}{b}\rfloor b$$ $$\frac{x+y+z}{b}-1<\lfloor \frac{x+y+z}{b}\rfloor $$

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