<< problem 62 - Cubic permutations Odd period square roots - problem 64 >>

# Problem 63: Powerful digit counts

The 5-digit number, 16807=7^5, is also a fifth power. Similarly, the 9-digit number, 134217728=8^9, is a ninth power.

How many n-digit positive integers exist which are also an nth power?

# Algorithm

check(n) finds all such powers with n digits:
- first it generates the smallest n-digit number from and the largest n-digit number to
- e.g. for n=5 we have from=10000 and to=99999
- then all numbers 1^n to 9^n are computed, if they are between from and to we have a match

While experimenting I saw no number with more than 21 digits to fulfilled the problem's conditions.

A minor headache was that 21 digits don't fit into C++ unsigned long long anymore.
Therefore I switched to double which has a few rounding issues but they don't affect the original problem.

## Alternative Approaches

It's easy to determine the number of digits using log10.
And as mentioned in the code, pow(a,b) returns a^b.
Both functions are available in the C++ standard library.

## Modifications by HackerRank

Hackerrank wants you to print all powers with a certain number of digits - instead of finding the number of all such powers.
However, the rounding issues of double now come into play and the right-most digits are a bit off for large n.

Fortunately, Hackerrank wants us only to find the powers with at most 19 digits ... and a 64 bit unsigned long long is sufficient for this task.
That's why you find the #ifdef construct where the type of Number is defined.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <iostream>

//#define ORIGINAL

// find all numbers where x^digits has digits digits (I'm loving that comment ...)
unsigned int check(unsigned int digits)
{
// numbe rof matches
unsigned int count = 0;

// unsigned long long isn't sufficient for the highest digits
// a double has some rounding issues but they don't affect to result
#ifdef ORIGINAL
typedef double Number;
#else
typedef unsigned long long Number;
#endif

// range of valid numbers
// from = 10^(digits-1)
// to   = 10^digits - 1
Number to = 1;
for (unsigned int i = 1; i <= digits; i++)
to *= 10;
Number from = to / 10;
to--;

// try all single-digit base numbers
for (unsigned int base = 1; base <= 9; base++)
{
// compute power = base ^ digits
Number power = base;
for (unsigned int i = 1; i < digits && power <= to; i++)
power *= base;
// could use C++'s pow(), too

// right number of digits ?
if (power >= from && power <= to)
{
count++;
#ifndef ORIGINAL
std::cout << std::fixed << power << std::endl;
#endif
}
}

return count;
}

int main()
{
#ifdef ORIGINAL
// check all digits
unsigned int count = 0;
for (unsigned int digits = 1; digits <= 21; digits++) // I observed no results with more than 21 digits
count += check(digits);
std::cout << count << std::endl;

#else

// check only certain digits
unsigned int digits = 9;
std::cin >> digits;
check(digits);
#endif

return 0;
}


This solution contains 11 empty lines, 14 comments and 9 preprocessor commands.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This live test is based on the Hackerrank problem.

Input data (separated by spaces or newlines):

This is equivalent to
echo 5 | ./63

Output:

(this interactive test is still under development, computations will be aborted after one second)

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

February 28, 2017 submitted solution

# Hackerrank

My code solves 10 out of 10 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

projecteuler.net/thread=63 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-63-n-digit-nth-power/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p063.java (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem063.go (written by Frederick Robinson)
Scala: github.com/samskivert/euler-scala/blob/master/Euler063.scala (written by Michael Bayne)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.
red problems are solved but exceed the time limit of one minute.

Please click on a problem's number to open my solution to that problem:

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The 206 solved problems had an average difficulty of 27.5% at Project Euler and
I scored 12,626 points (out of 14300 possible points, top rank was 20 out ouf ≈60000 in July 2017) at Hackerrank's Project Euler+.
Look at my progress and performance pages to get more details.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

 << problem 62 - Cubic permutations Odd period square roots - problem 64 >>
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