Project Euler Lab - Problem 698

#698 - 123 Numbers

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We define 123-numbers as follows:

  • 1 is the smallest 123-number.
  • When written in base 10 the only digits that can be present are "1", "2" and "3" and if present the number of times they each occur is also a 123-number.

So 2 is a 123-number, since it consists of one digit "2" and 1 is a 123-number. Therefore, 33 is a 123-number as well since it consists of two digits "3" and 2 is a 123-number.
On the other hand, 1111 is not a 123-number, since it contains 4 digits "1" and 4 is not a 123-number.

In ascending order, the first 123-numbers are:
\(1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 111, 112, 113, 121, 122, 123, 131, \ldots\)

Let \(F(n)\) be the \(n\)-th 123-number. For example \(F(4)=11\), \(F(10)=31\), \(F(40)=1112\), \(F(1000)=1223321\) and \(F(6000)= 2333333333323\).

Find \(F(111\,111\,111\,111\,222\,333)\). Give your answer modulo \(123\,123\,123\).

This problem is taken from Project Euler, Problem 698.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=698. Published Sunday, 19th January 2020, 07:00 am. Solved by 561 members at time of mirroring.

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