Question

Show that in any base $\mathbf{b}\mathbf{\ge }\mathbf{2}$, the sum of any three single-digit numbers is at most two digits long.

Short answer

Sum of any three single digit number is not as long as of a two-digit number.

Step-by-step solution

Find equation of sum of digits

Let’s assume $\mathrm{p},\mathrm{q},\mathrm{r}$ be the single digit numbers in base b.

Let sum of these number be$sum$given as:

$\mathit{s}\mathit{u}\mathit{m}\mathbf{=}\mathit{p}\mathbf{+}\mathit{q}\mathbf{+}\mathit{r}$

Given that:

Value of $\mathrm{p},\mathrm{q},\mathrm{r}$lies in range $\mathrm{b}\ge 2$

Here, $p,q,r$is at most $b-1$. For example: in base $10$ maximum single digit is $9$.

Calculate the value of sum

$\begin{array}{c}\mathrm{sum}=\mathrm{p}+\mathrm{q}+\mathrm{r}\\ =(\mathrm{b}-1)+(\mathrm{b}-1)+(\mathrm{b}-1)\\ =3(\mathrm{b}-1)\end{array}$

For $\mathrm{b}=2$,

$\begin{array}{c}\mathrm{sum}=3(2-1)\\ =3\times 1\\ =3\end{array}$

Here, sum is a single digit number.

For $\mathrm{b}=3$,

$\begin{array}{c}\mathrm{sum}=3(3-1)\\ =3\times 2\\ =6\end{array}$

Here, sum is a single digit number.

For $\mathrm{b}=9$,

$\begin{array}{c}\mathrm{sum}=3(9-1)\\ =3\times 8\\ =24\end{array}$

Here, sum is a two-digit number.

Thus, the sum is less than $b^2$ and $b^2$ in any base, sum is less than $b^2$ that is represented in two digits.