## A cyclic number with $n$ digits has a very interesting property:

When it is multiplied by $1, 2, 3, 4, \dots, n$, all the products have exactly the same digits, in the same order, but rotated in a circular fashion!

The smallest cyclic number is the $6$-digit number $142857$:

$142857 \times 1 = 142857$

$142857 \times 2 = 285714$

$142857 \times 3 = 428571$

$142857 \times 4 = 571428$

$142857 \times 5 = 714285$

$142857 \times 6 = 857142$

The next cyclic number is $0588235294117647$ with $16$ digits :

$0588235294117647 \times 1 = 0588235294117647$

$0588235294117647 \times 2 = 1176470588235294$

$0588235294117647 \times 3 = 1764705882352941$

$\dots$

$0588235294117647 \times 16 = 9411764705882352$

Note that for cyclic numbers, leading zeros are important.

There is only one cyclic number for which, the eleven leftmost digits are $00000000137$ and the five rightmost digits are $56789$ (i.e., it has the form $00000000137 \cdots 56789$ with an unknown number of digits in the middle). Find the sum of all its digits.

### The existence of cyclic numbers is intriguing, and it’s related to the decimal expansion of fractions whose denominator is a prime and does not divide the base. Here, we are dealing with base 10, so the primes that matter are those that aren’t divisors of 10 — that is 2 and 5.

For example, the repeating decimal 0.142857 corresponds to the fraction 1/7. The six-digit repeating block in this fraction corresponds to the $6$-digit cyclic number $142857$.

The length of the repeating block of a fraction with prime denominator $p$ is a factor of $p-1$. In the case of 1/7, $p-1 = 6$ which is the length of the repeating block $142857$. In the case of the second cyclic number given, the length of the repeating block is $16$ and the denominator is a prime number $p$ where $p – 1 = 16$, so the prime $p$ must be $17$.

In the question, the cyclic number given starts with $00000000137$ and ends with $56789$. We aren’t given the length of the repeating block, but we know it must be less than or equal to the length of the number, and must divide $p-1$ where $p$ is a (non-divisor of 10) prime number.

Let’s consider the number ends with $56789$, we know from the definition of cyclic numbers it’s a rotation of $p-1$ digits where p is a prime. So, it’s correctly can be associated with the prime number $67$, because $67-1=66$ and the length of cyclic number is $66$-digits.

This number is the cycle of $1/67$. To find the digits of it, we simply calculate the first 67 terms of the decimal cycle of $1/67$. We do long division and find that the quotient is:

“`

0001492537313432835820895522388059701492537313432835820895522388059701493

“`

As is evident from the decimal expansion of $1/67$, we simply take the repeating part as our cyclic number without considering the leading zero, which will be:

“`

1492537313432835820895522388059701492537313432835820895522388059701493

“`

Now, the problem asks for the sum of the digits. Due to the notion of cyclic numbers, each digit from 0 to 9 should occur $6$ times (66/10=6 remainder 6). We calculate the sum as follows:

$0*6+1*6+2*6+3*6+4*6+5*6+6*6+7*6+8*6+9*6 = 270$

So, the sum of the digits of the cyclic number is $270$.

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