Cyclic Numbers

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 :
$0588235294117647 \times 1 = 0588235294117647$
$0588235294117647 \times 2 = 1176470588235294$
$0588235294117647 \times 3 = 1764705882352941$
$0588235294117647 \times 16 = 9411764705882352$

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:


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:


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 $270$.

More Answers:
Maximal Coprime Subset
Largest Roots of Cubic Polynomials
Prime Generating Integers

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