<< problem 346 - Strong Repunits Sum of a square and a cube - problem 348 >>

# Problem 347: Largest integer divisible by two primes

The largest integer <= 100 that is only divisible by both the primes 2 and 3 is 96, as 96=32*3=25*3.
For two distinct primes p and q let M(p,q,N) be the largest positive integer <=N only divisible by both p and q
and M(p,q,N)=0 if such a positive integer does not exist.

E.g. M(2,3,100)=96.
M(3,5,100)=75 and not 90 because 90 is divisible by 2, 3 and 5.
Also M(2,73,100)=0 because there does not exist a positive integer <= 100 that is divisible by both 2 and 73.

Let S(N) be the sum of all distinct M(p,q,N). S(100)=2262.

Find S(10 000 000).

# My Algorithm

I go through all pairs (i,j) of primes i and j where i < j.
Then i analyze i^1: find the largest j^k_1 such that i^1 * j^k <= limit
In the second iteration I find the largest j^{k_2} such that i^2 * j^{k_2} <= limit
... and so on. Each time I check whether the product exceeds any previous product.

About 50% is comprised of the standard prime sieve from my toolbox.

# 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 100 | ./347

Output:

Note: the original problem's input 10000000 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>

// ---------- standard prime sieve from my toolbox ----------

// odd prime numbers are marked as "true" in a bitvector
std::vector<bool> sieve;

// return true, if x is a prime number
bool isPrime(unsigned int x)
{
// handle even numbers
if ((x & 1) == 0)
return x == 2;

// lookup for odd numbers
return sieve[x >> 1];
}

// find all prime numbers from 2 to size
void fillSieve(unsigned int size)
{
// store only odd numbers
const unsigned int half = size >> 1;

// allocate memory
sieve.resize(half, true);
// 1 is not a prime number
sieve[0] = false;

// process all relevant prime factors
for (unsigned int i = 1; 2*i*i < half; i++)
// do we have a prime factor ?
if (sieve[i])
{
// mark all its multiples as false
unsigned int current = 3*i+1;
while (current < half)
{
sieve[current] = false;
current += 2*i+1;
}
}
}

// ---------- problem-specific code ----------

int main()
{
unsigned long long limit = 10000000;
std::cin >> limit;

// compute all prime numbers
fillSieve(limit / 2);

// will contain the result
unsigned long long sum = 0;

// first prime
for (unsigned long long i = 2; i*i <= limit; i++)
{
// primes only
if (!isPrime(i))
continue;

// second prime
unsigned long long j = (i == 2) ? 3 : i + 2; // somewhat tricky way to express: "next odd number, start with 3"
for (; i*j <= limit; j += 2)
{
// primes only
if (!isPrime(j))
continue;

// largest number expressed as i^something * j^somethingelse (<= limit)
unsigned long long maxProduct = 0;

// note: j needs to be long long because otherwise a few overflows
//       lead to a result which is slightly off

// i^1 * j^1
auto product = i * j;
// for i^1, i^2, i^3, ... find the maximum exponent for j
do
{
// increase exponent of j as much as possible
auto current = product;
while (current * j <= limit)
current *= j;

// better than before ?
if (maxProduct < current)
maxProduct = current;

// increment i's exponent by one
product *= i;
} while (product <= limit);

sum += maxProduct;
}
}

// display result
std::cout << sum << std::endl;
return 0;
}


This solution contains 20 empty lines, 29 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.07 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 17, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 15% (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.

 << problem 346 - Strong Repunits Sum of a square and a cube - problem 348 >>
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