## Cyclic Numbers

Consider the number 2387. We can construct new numbers from the same digits by moving them cyclically:

3872, 8723, and 7238.

in general, we can construct n cyclic numbers from a number containing n digits:

$N_{1}=a_{1}a_{2}a_{3} \ldots a_{n-1}a_{n}$

$N_{2}=a_{2}a_{3} \ldots a_{n-1}a_{n}a_{1}$

$\vdots$

$N_{3}=a_{3}a_{4}a_{5}\ldots a_{n-1}a_{n}a_{1}a_{2}$

$N_{n}=a_{n}a_{1}a_{2}a_{3}\ldots a_{n-1}$.

What is the sum of all the cyclic numbers of such a set $\{N_{1}, N_{2}, \ldots, N_{n} \}$? it is easy to compute this from the above description. When we add all the numbers above, we find that in each vertical column, the sum simply is

$S = a_{1}+ a_{2} + a_{3} + \ldots + a_{n-1} + a_{n}$

so that if we factor it out, the remaining factor will be $111\ldots11$, with the number repeated n times. In our example, chosen above, we should get

$2387+3872+8723+7238=(2+3+8+7) \times 1111 = 20 \times 1111 = 22220$.

You can verify that this is indeed true.

There are several puzzles related to cyclic numbers. You are most welcome to share with us.

-Nalin Pithwa

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