<< problem 15 - Lattice paths | Number letter counts - problem 17 >> |

# Problem 16: Power digit sum

(see projecteuler.net/problem=16)

2^{15} = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.

What is the sum of the digits of the number 2^{1000}?

# Algorithm

These two equations describe the iterative computation of 2^x:

2^0 = 1 and 2^x = 2 * 2^{x-1}

These numbers grow pretty big and for x=64 exceed the range of `unsigned long long`

.

That's why I store all decimal `Digits`

in a `std::vector`

, where the lowest index contains the least significant digits.

For x=0 - which represents 2^0 = 1 - this `Digits`

container's elements are `{ 1 }`

For x=15 - which represents 2^{15} = 1 - this `Digits`

container's elements are `{ 8, 6, 7, 2, 3 }`

Multiplying `Digits`

by 2 follows the same rules as basic multiplication taught in school:

1. multiply each digit by 2, start at the lowest digit

2. if digit * 2 >= 10 then an overflow occurred: carry over (digit * 2) \div 10 to the next digit and keep (digit * 2) \mod 10.

*Note:* we must carry over at most 1 and instead of a computationally expensive modulo we can subtract 10 which is faster

## Alternative Approaches

Exponentation by squaring saves many steps:

The result of 2^1024 can be found in just 10 steps instead of 1024.

## Modifications by HackerRank

To avoid timeouts, I store all results (even intermediate steps) in a cache and re-use as much as possible from this cache.

# My code

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

#include <vector>
#include <iostream>
// store single digits in an array, lowest digit come first

typedef std::vector<unsigned int> Digits;
int main()
{
// memoize powers of two
std::vector<Digits> cache;
// add 2^0 = 1
cache.push_back({ 1 });
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int exponent;
std::cin >> exponent;
// and compute the remaining exponents
for (unsigned int current = cache.size(); current <= exponent; current++)
{
auto power = cache.back();
unsigned int carry = 0;
for (auto& i : power)
{
// times two ...
i = 2 * i + carry;
// handle overflow
if (i >= 10)
{
i -= 10;
carry = 1;
}
else
{
carry = 0;
}
}
// still some carry left ?
if (carry != 0)
power.push_back(carry);
// memoize result
cache.push_back(power);
}
// sum of all digits
unsigned int sum = 0;
for (auto i : cache[exponent])
sum += i;
std::cout << sum << std::endl;
}
return 0;
}

This solution contains 9 empty lines, 9 comments and 2 preprocessor commands.

# Interactive test

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

This is equivalent to`echo "1 15" | ./16`

Output:

*Note:* the original problem's input `1000`

__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 **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 23, 2017 submitted solution

March 31, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler016

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.

# Similar problems at Project Euler

Problem 20: Factorial digit sum

*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=16 - **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-16/ (written by Kristian Edlund)

Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p016.hs (written by Nayuki)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p016.java (written by Nayuki)

Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p016.mathematica (written by Nayuki)

C: github.com/eagletmt/project-euler-c/blob/master/10-19/problem16.c (written by eagletmt)

Go: github.com/frrad/project-euler/blob/master/golang/Problem016.go (written by Frederick Robinson)

Javascript: github.com/dsernst/ProjectEuler/blob/master/16 Power digit sum.js (written by David Ernst)

Scala: github.com/samskivert/euler-scala/blob/master/Euler016.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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