<< problem 277 - A Modified Collatz sequence | Scoring probabilities - problem 286 >> |

# Problem 278: Linear Combinations of Semiprimes

(see projecteuler.net/problem=278)

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)

# My code

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

#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.

# 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 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)*

# Benchmark

The correct solution to the original Project Euler problem was found in 0.11 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

July 11, 2017 submitted solution

July 11, 2017 added comments

# Difficulty

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

# Links

projecteuler.net/thread=278 - **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.

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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 277 - A Modified Collatz sequence | Scoring probabilities - problem 286 >> |