Project Euler Lab - Problem 686

#686 - Powers of Two

● AppliedOfficial difficulty: 14%PolynomialsTier B - browser, with the efficient algorithmNot viewed
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\(2^7=128\) is the first power of two whose leading digits are "12".
The next power of two whose leading digits are "12" is \(2^{80}\).

Define \(p(L, n)\) to be the \(n\)th-smallest value of \(j\) such that the base 10 representation of \(2^j\) begins with the digits of \(L\).
So \(p(12, 1) = 7\) and \(p(12, 2) = 80\).

You are also given that \(p(123, 45) = 12710\).

Find \(p(123, 678910)\).

This problem is taken from Project Euler, Problem 686.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=686. Published Saturday, 26th October 2019, 07:00 pm. Solved by 4,297 members at time of mirroring.

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Prerequisites

Lessons that prepare you:
1.1 Sets, Functions, and Relations · 19.10 Search: Backtracking, Branch-and-Bound, Binary Search, Meet-in-the-Middle · 2.1 Functions, Limits, and Continuity · 4.2 Linear Maps, Matrices, Rank, and the Null Space

Recommended stepping-stone problems: #63 · #71 · #112

Concepts: algebra

Likely techniques: binary-search

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