<< problem 181 - Investigating in how many ways objects of two ... | Number Mind - problem 185 >> |

# Problem 183: Maximum product of parts

(see projecteuler.net/problem=183)

Let N be a positive integer and let N be split into k equal parts, r = N/k, so that N = r + r + ... + r.

Let P be the product of these parts, P = r * r * ... * r = r^k.

For example, if 11 is split into five equal parts, 11 = 2.2 + 2.2 + 2.2 + 2.2 + 2.2, then P = 2.2^5 = 51.53632.

Let M(N) = P_max for a given value of N.

It turns out that the maximum for N = 11 is found by splitting eleven into four equal parts which leads to P_max = (11/4)^4;

that is, M(11) = 14641/256 = 57.19140625, which is a terminating decimal.

However, for N = 8 the maximum is achieved by splitting it into three equal parts, so M(8) = 512/27, which is a non-terminating decimal.

Let D(N) = N if M(N) is a non-terminating decimal and D(N) = -N if M(N) is a terminating decimal.

For example, sum{D(N)} for 5 <= N <= 100 is 2438.

Find sum{D(N)} for 5 <= N <= 10000.

# My Algorithm

My first approach was to have a simple loop running from k = 1 to n:

- find the largest (n/k)^k

- figure out whether (n/k)^k terminates

When I looked at the values of k I saw that n/k was always approx 2.7.

And Wolfram Alpha confirmed my suspicion: www.wolframalpha.com/input/?i=max((8%2Fn)%5En)

M(8) = max((8/n)^n) = e^{8/e} at k = 8/e where e = 2.7182818284..., see en.wikipedia.org/wiki/E_(mathematical_constant)

In general: max((x/n)^n) = e^{n/e} at k = n/e

Unfortunately, n/e isn't an integer and therefore the best k is `k = round(n / 2.718281828)`

(that's enough digits of e for this problem)

The number (n/k)^k terminates if n/k terminates.

And n/k terminates if the greatest common divisor gcd(n, k) has only 2 and 5 as prime factors (something I learnt in problem 26).

## Alternative Approaches

As mentioned before, my initial approach computed all (n/k)^k.

But numbers became huge very fast ... well, I didn't need their "true" value: it's sufficient to have their logarithm because if a < b then log(a) < log(b), too.

In the end I converted (x/n)^n to n log(x/n).

Using `round`

can be dangerous if not enough digits of e are used. 10 digits are totally fine for that problem, though.

## Modifications by HackerRank

I need to cache intermediate result to be able to handle the huge number of input values.

# My code

… was written in C++ 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>
#include <vector>
#include <cmath>
int main()
{
// five dummy entries
std::vector<long long> cache(5, 0);
#define ORIGINAL
#ifndef ORIGINAL
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
#endif
{
unsigned int limit = 10000;
std::cin >> limit;
auto result = cache.back();
for (unsigned int n = cache.size(); n <= limit; n++)
{
unsigned int k = round(n / 2.718281828);
// terminating only if n/k has only prime factors 2 and 5
while (k % 5 == 0)
k /= 5;
while (k % 2 == 0)
k /= 2;
if (n % k == 0)
result -= n;
else
result += n;
// cache intermediate results, too (needed for Hackerrank only)
cache.push_back(result);
}
// display result
std::cout << cache[limit] << std::endl;
}
return 0;
}

This solution contains 8 empty lines, 4 comments and 6 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 100 | ./183`

Output:

*Note:* the original problem's input `10000`

__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

June 13, 2017 submitted solution

June 13, 2017 added comments

July 28, 2017 modified to solve Hackerrank, too

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler183

My code solves **11** out of **11** test cases (score: **100%**)

# Difficulty

Project Euler ranks this problem at **45%** (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.

# Links

projecteuler.net/thread=183 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

# 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 or the memory limit of 256 MByte.

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

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I scored 13,183 points (out of 15300 possible points, top rank was 17 out of ≈60000 in August 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.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.

Thanks for all their endless effort.

<< problem 181 - Investigating in how many ways objects of two ... | Number Mind - problem 185 >> |