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

# Problem 63: Powerful digit counts

(see projecteuler.net/problem=63)

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 `#ifdef`

s 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.

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

April 26, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler063

My code solved **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 never an option.

# Links

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 already 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.

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

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