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I noticed that, starting with 0.5 you can divide by 2 twice by inserting a number in the previous result after the decimal point. Specifically, to go from 0.5 to 0.25 a number 2 in inserted after the decimal point and to go from 0.25 to 0.125 a number 1 is inserted after the decimal point.

0.  5
0. 25
0.125

Is this the longest sequence where this happens?

If we chose the base of the number as divisor, then 0.1 (in that base) generates an infinite sequence (0.1, 0.01, 0.001, etc...). But what about base $B$ with divisor $D\ne B$?

Miguel
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1 Answers1

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Let $x$ be the number we're dividing by $D$, and $k$ the digit we add, in base $b$. Then $$\frac{x}{b} + \frac{k}{b} = \frac{x}{D}$$ so $x = \frac{D k}{b-D}$.

Since $k \in \{0,1,2,\dots, b\}$, have $$x \in \{0,\frac{D}{b-D},2 \times \frac{D}{b-D},\dots,(b-1) \times \frac{D}{b-D}\}$$

These are the only numbers which we may divide by $D$ in the way you want. We wish to find a sequence of longest length in this, which means we want $b-1$ to be a power of $D$. That makes the longest possible sequence.

Patrick Stevens
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  • Hi Patrick, thank you for your interesting answer. I'm having trouble fully understanding it though. First, if I plug b=10 and D=2 (original problem) I get x in {0, 0.25, 0.5, 0.75, 1.0, 1.25, 1.50, 1.75, 2.00, 2.25}. This solves your equation but doesn't work for numbers greater or equal to 1. – Miguel Oct 08 '15 at 22:06
  • Also @Patrick, could you explain the last two sentences? I tried to apply it to b=10, D=3 but I didn't understand how it would work. – Miguel Oct 13 '15 at 22:08