# Problem 278: Linear Combinations of Semiprimes

Given the values of integers 1 < a_1 < a_2 < ... < a_n, consider the linear combination
q_1 a_1 + q_2 a_2 + ... + q_n a_n = b, using only integer values q_k >= 0.

Note that for a given set of a_k, it may be that not all values of b are possible.
For instance, if a_1 = 5 and a_2 = 7, there are no q_1 >= 0 and q_2 >= 0 such that b could be
1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18 or 23.
In fact, 23 is the largest impossible value of b for a_1 = 5 and a_2 = 7.
We therefore call f(5, 7) = 23.
Similarly, it can be shown that f(6, 10, 15) = 29 and f(14, 22, 77) = 195.

Find sum{f(pq,pr,qr)}, where p, q and r are prime numbers and p < q < r < 5000.

# My Algorithm

Assume I have two numbers x and y where gcd(x,y)=1.
The value m = xy - x - y can't be represented with some coefficients m = px + qy because:
xy - x - y = px + qy
xy = px + qy + x + y
xy = (p+1)x + (q+1)y

xy is a multiple of x and (p+1)x is a multiple of x, hence (q+1)y should be a multiple of x, too.
xy is a multiple of y and (q+1)y is a multiple of y, hence (p+1)x should be a multiple of y, too.
But gcd(x,y)=1 so y can't be a multiple of x and therefore q+1 should be a multiple of x.
And for the same reason x can't be a multiple of y and therefore p+1 should be a multiple of y.
Possible values for q+1 would be x, 2x, 3x, ... (and for p+1: y, 2y, 3y, ...)
If I assume the lowest value p+1=y and q+1=x then the equation becomes
xy = y * x + x * y
xy = 2xy → contradition !

Therefore m = xy - x - y actually can't be represented with some coefficients m = px + qy.

With three numbers x,y,z and gcd(x,y,z)=1 the idea is very similar:
if there would be some coefficients p, q and r such that m = pxy + qxz + ryz represents m = 2xyz - xy - xz - yz then
(2xyz - xy - xz - yz) mod x = -yz
pxy + qxz + ryz = 2xyz - xy - xz - yz
2xyz = pxy + qxz + ryz + xy + xz + yz
2xyz = (py+qz+y+z)x + (rz + z)y
Hence rz + z = (r+1)z must be a multiple of x. z can't be such a multiple (because of gcd(x,y,z) = 1).
The same idea for y and z gives that p+1 must be a multiple of z and q+1 a multiple of y.
As before - if I choose the smallest possible p+1=z, q+1=y and r+1=x:
2xyz = zxy + yxz + xyz + xy + xz + yz
2xyz = 3xyz + xy + xz + yz → contradiction

I didn't come up with the full solution, I just know how to use search engines :-;
I found the problem in the 24th International Mathemtical Olympiad held 1983 in Paris, France
somewhat cryptic solution: www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln833.html
I stumbled across it while reading the German Wikipedia de.wikipedia.org/wiki/Münzproblem
unfortunately, the English page misses that special case en.wikipedia.org/wiki/Coin_problem
but it can be derived from their n=2 explanations (pretty much what I have done above)

# 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 10 | ./278

Output:

Note: the original problem's input 5000 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)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       #include <iostream>
#include <vector>

int main()
{
unsigned int limit = 5000;
std::cin >> limit;

// simple prime sieve from my toolbox
std::vector<unsigned long long> primes = { 2 };
for (unsigned int i = 3; i <= limit; i += 2)
{
bool isPrime = true;

// test against all prime numbers we have so far (in ascending order)
for (auto x : primes)
{
// prime is too large to be a divisor
if (x*x > i)
break;

// divisible => not prime
if (i % x == 0)
{
isPrime = false;
break;
}
}

// yes, we have a prime
if (isPrime)
primes.push_back(i);
}

// all combinations of primes
unsigned long long sum = 0;
for (size_t i = 0; i < primes.size(); i++)
for (size_t j = i + 1; j < primes.size(); j++)
for (size_t k = j + 1; k < primes.size(); k++)
{
auto p = primes[i];
auto q = primes[j];
auto r = primes[k];
sum += 2*p*q*r - p*q - p*r - q*r;
}

std::cout << sum << std::endl;
return 0;
}


This solution contains 7 empty lines, 6 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.11 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++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

July 11, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 50% (out of 100%).

# Heatmap

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

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

more about me can be found on my homepage, especially in my coding blog.
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