# Numbers that have constant digits value independent of base

I was wondering why for example $\dfrac{1111_b \cdot 111_b}{11_b \cdot 1_b} = 11211_b$. Is there a good explanation for this and is there a name for products like this which have constant digits value independent of base? It also seems to hold with $\dfrac{11111_b \cdot 1111_b \cdot 111_b}{111_b \cdot 11_b \cdot 1_b}$ and other numbers as well.

Here is my general question:

Prove that $$R(m,k) = \frac{ 1^{(m)}_b \cdot 1^{(m-1)}_b\cdots 1^{(m-k+1)}_b}{1^{(1)}_b \cdot 1^{(2)}_b\cdots 1^{(k)}_b} = c_b$$ where $c$ is a constant and $b>2$.

• The digit 2 does not exist in binary. – mathreadler Jun 25 '16 at 13:02
• @mathreadler For any base other than $2$. – Puzzled417 Jun 25 '16 at 13:03
• @Puzzled417, Write $$1111_b$$ as $$1+b+b^2+b^3$$ etc. – lab bhattacharjee Jun 25 '16 at 13:05
• Typo in the numerator with $111_b$? – Piquito Jun 25 '16 at 13:17

For example the first one: $1111_b$ can first be divided by $11_b$ to produce $101_b$, then the multiplication will be $111_b + 100_b\cdot 111_b = 11211_b$
• In fact, the generalization is in terms of repunits. Let $$R(m,k) = \frac{ 1^{(m)}_b \cdot 1^{(m-1)}_b\cdots 1^{(m-k+1)}_b}{1^{(1)}_b \cdot 1^{(2)}_b\cdots 1^{(k)}_b} = c_b$$ where $c$ is a constant. – Puzzled417 Jun 25 '16 at 13:14