Project Euler Lab - Problem 358

#358 - Cyclic Numbers

● AdvancedOfficial difficulty: 49%Naive enumeration is infeasible; requires a mathematical reductionTier C - reduced scale in browser; full scale in notebookNot viewed
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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.

This problem is taken from Project Euler, Problem 358.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=358. Published Saturday, 12th November 2011, 07:00 pm. Solved by 1,877 members at time of mirroring.

Why this is useful

Optimization. The transferable skill is replacing infeasible enumeration with a mathematical reduction - the core move in calibration and large-scale computation (Phases 10, 13).

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Prerequisites

Lessons that prepare you:
19.14 Computational Complexity, Feasibility Estimation, and Proving Algorithms Correct · 19.7 Dynamic Programming: Memoization and Tabulation

Recommended stepping-stone problems: #637 · #749 · #318

Concepts: brute-force-reduction

Likely techniques: digit-dp

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