<< problem 29 - Distinct powers Coin sums - problem 31 >>

# Problem 30: Digit fifth powers

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

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.

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

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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+.
 << problem 29 - Distinct powers Coin sums - problem 31 >>
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