Number Rotations

Consider the number $142857$. We can right-rotate this number by moving the last digit ($7$) to the front of it, giving us $714285$.
It can be verified that $714285 = 5 \times 142857$.
This demonstrates an unusual property of $142857$: it is a divisor of its right-rotation.
Find the last $5$ digits of the sum of all integers $n$, $10 \lt n \lt 10^{100}$, that have this property.

This is a well-known problem in number theory and requires some mathematical tricks and logical reasoning. The problem can be broken down into two parts: identifying numbers that possess that rotational property, and then finding the last five digits of the sum of all such numbers.

First, let’s inspect “142857”. We can notice that every right rotation gives us a multiple of our original number (142857).
Right rotating once gives: 714285 = 5 * 142857.
Based on this property, we can list down right rotations of “142857”:

1 * 142857 = 142857 ;
10 * 142857 = 1428570 ;
100 * 142857 = 14285700 ;
1000 * 142857 = 142857000 ;
10000 * 142857 = 1428570000 ;
100000 * 142857 = 14285700000.

In all of these cases, moving the last digit to the front gives us a multiple of the original number. Interesting to note here is the relationship between the multiplier and the power of 10 just before the number.

Now, let’s consider the number itself. “142857” is a repeating decimal of 1/7, that is evident from the fact it’s multiplied with 1, 10, 100, 1000, 10000, 100000 to get the rotated versions. This suggests that it is not the only number with this property and there may be other fractions which repeating decimals have this property.

After some mathematical exploration, one can find that only fractions of the form 1/(2^a × 5^b) where a,b are non-negative integers give decimals with the rotational property. This is because rotations are in essence multiplying by powers of 10 and only these denominators are divisible by 10.

So, we need to find numbers of form k*(10^n – 1)/9 where “k” and “n” are integers. Also, due to the requirement 10 < n < 10^100, also n ranges from 2 to 100. Now, the second part of the problem suggests finding the sum of all those numbers which satisfy the conditions above. Also, it asks us to determine just the last five digits of that sum. To do this, we should take every sum modulo 100000 (because we are interested only in the last five digits). The complete Python program below solves this problem using the reasoning discovered above: ```python # Function to solve the problem def rotational_divisor_sum(limit): modulus = 100000 total = 0 power_of_10_mod = 10 # Iterating for all n from 2 to limit (here 100) for _ in range(2, limit + 1): total += (power_of_10_mod - 1) // 9 total %= modulus power_of_10_mod *= 10 power_of_10_mod %= modulus return total # Solving for numbers under the limit 10^100 print(rotational_divisor_sum(100)) ``` This will result in the last 5 digits of the described sum. Please replace "limit" in the rotational_divisor_sum function with the required value to compute different sums.

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