Problem 30: Digit fifth powers

(see projecteuler.net/problem=30)

Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:

1634 = 1^4 + 6^4 + 3^4 + 4^4
8208 = 8^4 + 2^4 + 0^4 + 8^4
9474 = 9^4 + 4^4 + 7^4 + 4^4

As 1 = 1^4 is not a sum it is not included.

The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

Algorithm

The sum of the digits' fifth powers is maximized when each digit is 9:
1 digit: 1 * 9^5 = 59049
2 digits: 2 * 9^5 = 118098
3 digits: 3 * 9^5 = 177147
4 digits: 4 * 9^5 = 236196
5 digits: 5 * 9^5 = 295245
6 digits: 6 * 9^5 = 354294
7 digits: 7 * 9^5 = 413343

The last line is pretty interesting: it's impossible for a seven-digit number to have a seven-digit sum of its digits' fifth powers,
because all those sums would have at most six digits.

If we analyse all numbers from 2 to 354294 (maximum sum for 6 digits) then we can solve the problem:
1. split each number into its digits
2. add all fifth powers of these digits
3. if the sum is equal to the original number then add it to our result

Modifications by HackerRank

The exponent varies between 3 and 6. The maximum number can be computed similar to the table above:
7 digits: 7 * 9^6 = 3720087
8 digits: 8 * 9^6 = 4251528 → impossible

To simplify the code my loop always stops at 4251528 even though lower numbers would suffice for exponents < 6.
Timeouts are no issue here.

My code

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

#include <iostream>
 
int main()
{
unsigned int exponent;
std::cin >> exponent;
 
// result
unsigned int sum = 0;
 
// there can't be a number with 8 digits (or more) which fulfils the condition for exponent=6
// if all digits are 9s, then
// 7 digits: 7 * 9^6 = 3720087
// 8 digits: 8 * 9^6 = 4251528
for (unsigned int i = 2; i <= 7*9*9*9*9*9*9; i++)
{
// sum of i's digits to the power of "exponent"
unsigned int thisSum = 0;
 
// split current number into its digit
unsigned int reduce = i;
while (reduce > 0)
{
// take the right-most digit
unsigned int digit = reduce % 10;
// and remove it
reduce /= 10;
 
// compute digit^exponent, could use pow() as well
unsigned int power = 1;
for (unsigned int j = 1; j <= exponent; j++)
power *= digit;
 
// add to thisSum
thisSum += power;
}
 
// sum of digits^n equal to the original number ?
if (thisSum == i)
sum += i;
}
 
// and we're done
std::cout << sum << std::endl;
return 0;
}

This solution contains 8 empty lines, 13 comments and 1 preprocessor command.

Interactive test

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

Input data (separated by spaces or newlines):

This is equivalent to
echo 4 | ./30

Output:

(please click 'Go !')

Note: the original problem's input 5 cannot be entered
because just copying results is a soft skill reserved for idiots.

(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 0.15 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 23, 2017 submitted solution
April 6, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler030

My code solved 4 out of 4 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.

Similar problems at Project Euler

Problem 34: Digit factorials

Note: I'm not even close to solving all problems at Project Euler. Chances are that similar problems do exist and I just haven't looked at them.

Links

projecteuler.net/thread=30 - 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-30-sum-numbers-that-can-be-written-as-the-sum-fifth-powers-digits/ (written by Kristian Edlund)
Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p030.hs (written by Nayuki)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p030.java (written by Nayuki)
Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p030.mathematica (written by Nayuki)
C: github.com/eagletmt/project-euler-c/blob/master/30-39/problem30.c (written by eagletmt)
Go: github.com/frrad/project-euler/blob/master/golang/Problem030.go (written by Frederick Robinson)
Javascript: github.com/dsernst/ProjectEuler/blob/master/30 Digit fifth powers.js (written by David Ernst)
Scala: github.com/samskivert/euler-scala/blob/master/Euler030.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.

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The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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