# Cyclic shifts when multiplied by $2$.

I was trying to solve the following problem:

Find a number in base $10$, which when multiplied by $2$, results in a number which is a cyclic shift of the original number, such that the last digit (least significant digit) becomes the first digit.

I believe one such number is $105263157894736842$

This I was able to get by assuming the last digit and figuring out the others, till there was a repeat which gave a valid number.

I noticed that each time (with different guesses for the last digit), I got an $18$ digit number which was a cyclic shift of the one I got above.

Is there any mathematical explanation for the different answers being cyclic shifts of each other?

• The short answer is that these numbers are $\frac{n}{19}$ expanded out. – Calvin Lin May 15 '13 at 14:26

Suppose an $N$ digit number satisfies your condition, write it as $N= 10a + b$, where $b$ is the last digit. Then, your condition implies that

$$2 (10a + b) = b\times 10^{N-1} + a$$

Or that $b \times (10^{N-1} -1 ) = 19 a$.

Clearly, $b$ is not a multiple of 19, so we must have $10^{N-1} -1$ to be a multiple of 19. You should be able to verify (modular arithmetic), that this happens if and only if $N\equiv 0 \pmod{19}$. This gives us that $N = 19k$, and the numbers have the form

$$\frac{ 10^{19k} + 9 } {19} b = 10 \lfloor \frac{10^{19k-1} } {19} b \rfloor$$

[The equality occurs because $\frac{1}{19} < 1$.]

I now refer you to "When is $\frac{1}{p}$ obtained as a cyclic shift", for you to reach your conclusion.

There is one exception, where we need the leading digit to be considered as 0 (not the typical definition, so I though I'd point it out). We can get more solutions, obtained by concatenations of your number with itself, and several cyclic shifts (and again accounting for a possible leading digit of 0).