<< problem 563 - Robot Welders Super Pandigital Numbers - problem 571 >>

# Problem 565: Divisibility of sum of divisors

Let sigma(n) be the sum of the divisors of n.
E.g. the divisors of 4 are 1, 2 and 4, so sigma(4)=7.

The numbers n not exceeding 20 such that 7 divides sigma(n) are: 4,12,13 and 20, the sum of these numbers being 49.

Let S(n,d) be the sum of the numbers i not exceeding n such that d divides sigma(i).
So S(20,7)=49.

You are given: S(10^6,2017)=150850429 and S(10^9,2017)=249652238344557

Find S(10^11,2017).

# My Algorithm

(1) \sigma_1(p) = p + 1 if p is prime

(2) \sigma_1(p^k) = dfrac{p^{k+1} - 1}{p - 1} if p is prime

(3) \sigma_1(m * n) = \sigma_1(m) * \sigma_1(n) if m and n are coprime

The most crucial observation is that if \sigma_1(m) is divisible by 2017, then according to (3) almost every multiple is, too
(the exception is when m and n are not coprime, for example for m = n).

My function search() starts with equation (2) (and k >= 2) because there are only few such numbers.
I check for every prime whether \sigma_1(p^k) = dfrac{p^{k+1} - 1}{p - 1} is a multiple of 2017.
If yes, then that number including all its multiples (except every p-th multiple) has a divisor sum that is divisible by 2017.
Since p^2 <= 10^11 less than a million number have to be checked which is done in less than a second.

Much more computation time is spent on equation (1):
every prime p where p+1 is a multiple of 2017 and its multiples have a matching divisor sum.
Instead of finding all primes below 10^11 I check every number 2017k - 1 whether it is prime.
The Miller-Rabin test from my toolbox is pretty fast but I had to include a GCC optimization to stay below one minute execution time.

A substantial part of my code is devoted to detecting duplicates: a few number are found multiple times.
For example 12101 is the smallest prime where \sigma_1 is divisible by 2017. The next prime is 24203.
Their product 12101 * 24203 = 292880503 will be found while processing all multiples of 12101 and then again while processing all multiples of 24203.
The container found is a simple std::vector because it has pretty much zero data structure overhead (unlike std::set).

Duplicates (or a "collision") is impossible for the first 12101-1=12100 multiples of a matching prime (since 12101 is the smallest).
To keep my code more generic I simplified that idea and only assume that a collision is impossible for the first 2017-1=2016 multiple.
If the current prime p is larger than sqrt{10^11} then a collision can only occur if the multiple was seen before because it must be smaller than p.
Therefore I sort the list of primes found so far and perform a lookup with binary search.
All these optimizations have only one goal: reduce the size of found until it drops below the 256 MByte limit.

When all primes are processed then found is sorted again to eliminate all duplicates with std::unique.

## Note

I had the solution pretty fast but my code needed about 6 minutes and almost 2 GByte RAM.
Most of the time was spent optimizing the code (e.g. refactor from std::set to std::vector).
And while optimizing, I "broke" the algorithm quite a few times ...

Right now my solution has still high memory consumption and execution time.
It's by far the highest ranking expensive solution which doesn't exceed the memory or CPU limits.

# 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):
Note: The second parameter must be a prime else the result will be garbage

This is equivalent to
echo "1000000 2017" | ./565

Output:

Note: the original problem's input 100000000000 2017 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>
#include <algorithm>
#include <cmath>

// ---------- mulmod, powmod and Miller-Rabin test from my toolbox ----------

// return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;

// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

#ifdef __GNUC__
// based on GCC's 128 bit implementation
return ((unsigned __int128)a * b) % modulo;
#endif

// we might encounter overflows (slow path)
// the number of loops depends on b, therefore try to minimize b
if (b > a)
std::swap(a, b);

// bitwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
// b is odd ? a*b = a + a*(b-1)
if (b & 1)
{
result += a;
if (result >= modulo)
result -= modulo;
// skip b-- because the bit-shift at the end will remove the lowest bit anyway
}

// b is even ? a*b = (2*a)*(b/2)
a <<= 1;
if (a >= modulo)
a -= modulo;

// next bit
b >>= 1;
}

return result;
}

// return (base^exponent) % modulo => simple implementation
unsigned long long powmod(unsigned long long base, unsigned long long exponent, unsigned long long modulo)
{
unsigned long long result = 1;
while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
result = mulmod(result, base, modulo);

// even exponent ? a^b = (a*a)^(b/2)
base = mulmod(base, base, modulo);
exponent >>= 1;
}
return result;
}

// Miller-Rabin-test
bool isPrime(unsigned long long p)
{
// IMPORTANT: requires mulmod(a, b, modulo) and powmod(base, exponent, modulo)

// some code from             https://ronzii.wordpress.com/2012/03/04/miller-rabin-primality-test/
// with optimizations from    http://ceur-ws.org/Vol-1326/020-Forisek.pdf
// good bases can be found at http://miller-rabin.appspot.com/

// trivial cases
const unsigned int bitmaskPrimes2to31 = (1 <<  2) | (1 <<  3) | (1 <<  5) | (1 <<  7) |
(1 << 11) | (1 << 13) | (1 << 17) | (1 << 19) |
(1 << 23) | (1 << 29); // = 0x208A28Ac
if (p < 31)
return (bitmaskPrimes2to31 & (1 << p)) != 0;

if (p %  2 == 0 || p %  3 == 0 || p %  5 == 0 || p % 7 == 0 || // divisible by a small prime
p % 11 == 0 || p % 13 == 0 || p % 17 == 0)
return false;

if (p < 17 * 19) // we filtered all composite numbers < 17*19, all others below 17*19 must be prime
return true;

// test p against those numbers ("witnesses")
// good bases can be found at http://miller-rabin.appspot.com/
const unsigned int STOP = 0;
const unsigned int TestAgainst1[] = { 377687, STOP };
const unsigned int TestAgainst2[] = { 31, 73, STOP };
const unsigned int TestAgainst3[] = { 2, 7, 61, STOP };
// first three sequences are good up to 2^32
const unsigned int TestAgainst4[] = { 2, 13, 23, 1662803, STOP };
const unsigned int TestAgainst7[] = { 2, 325, 9375, 28178, 450775, 9780504, 1795265022, STOP };

// good up to 2^64
const unsigned int* testAgainst = TestAgainst7;
// use less tests if feasible
if (p < 5329)
testAgainst = TestAgainst1;
else if (p < 9080191)
testAgainst = TestAgainst2;
else if (p < 4759123141ULL)
testAgainst = TestAgainst3;
else if (p < 1122004669633ULL)
testAgainst = TestAgainst4;

// find p - 1 = d * 2^j
auto d = p - 1;
d >>= 1;
unsigned int shift = 0;
while ((d & 1) == 0)
{
shift++;
d >>= 1;
}

// test p against all bases
do
{
auto x = powmod(*testAgainst++, d, p);
// is test^d % p == 1 or -1 ?
if (x == 1 || x == p - 1)
continue;

// now either prime or a strong pseudo-prime
// check test^(d*2^r) for 0 <= r < shift
bool maybePrime = false;
for (unsigned int r = 0; r < shift; r++)
{
// x = x^2 % p
// (initial x was test^d)
x = mulmod(x, x, p);
// x % p == 1 => not prime
if (x == 1)
return false;

// x % p == -1 => prime or an even stronger pseudo-prime
if (x == p - 1)
{
// next iteration
maybePrime = true;
break;
}
}

// not prime
if (!maybePrime)
return false;
} while (*testAgainst != STOP);

// prime
return true;
}

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

// much faster ... but still takes a few seconds
unsigned long long search(unsigned long long limit, unsigned int multiple)
{
std::vector<unsigned long long> found;
// avoid re-allocations for limit = 10^11
if (limit == 100000000000ULL)
found.reserve(26240000);

// find primes where sigma(prime^2) is divisible by 2017 (same for sigma(prime^3), ...)
for (unsigned long long p = 2; p*p <= limit; p++)
{
if (!isPrime(p))
continue;

// at least p^2
auto power = p * p;
while (power <= limit)
{
// sigma = (p^q - 1) / (p - 1)
auto sigma = (power*p - 1) / (p - 1);
if (sigma % multiple == 0)
// including all multiples
for (auto i = 1; i*power <= limit; i++)
if (i % p != 0) // avoid the case where i is a multiple of p because then they aren't coprime anymore
found.push_back(i*power);

// next iteration will exceed limit ?
if (limit / power < p)
break;

// keep going ...
power *= p;
}
}

// sum of all matches
unsigned long long result = 0;

// switch to "optimized" mode if p^2 > limit
size_t sortedSize = 0;

// sigma(prime) = prime + 1
// => check all numbers 2017k - 1 whether they are prime
for (unsigned long long p = multiple - 1; p <= limit; p += multiple)
{
// only primes ...
if (!isPrime(p))
continue;

// "optimized" mode:
// collisions can only occur when multiplying p by a smaller prime which must be part of the "found" container
if (sortedSize == 0 && p*p > limit)
{
// sort all solutions so that I can search for elements with std::binary_search
std::sort(found.begin(), found.end());
sortedSize = found.size();
}

// sigma(p) is a multiple of 2017 and that's also true for all its multiples
for (unsigned long long i = 1; i*p <= limit; i++)
{
// avoid the case where i is a multiple of p because then they aren't coprime anymore
if (i % p == 0)
continue;

// another solution (which might be a duplicate, though !)
auto current = i * p;

// if i is small then a collision is impossible
if (i < multiple - 1)
{
result += current;
continue;
}

// I can detect collisions immediately if p > sqrt(limit) because then p > i
if (sortedSize > 0)
{
if (!std::binary_search(found.begin(), found.begin() + sortedSize, current))
result += current;
}
else
// potential collision
found.push_back(current);
}
}

// exclude duplicates
std::sort(found.begin(), found.end());
auto duplicates = std::unique(found.begin(), found.end());
// sum of all found numbers (but count duplicates only once)
for (auto i = found.begin(); i != duplicates; i++)
result += *i;

return result;
}

int main()
{
unsigned long long limit = 100000000000ULL; // 10^11
unsigned int    multiple = 2017;
std::cin  >> limit >> multiple;
std::cout << search(limit, multiple) << std::endl;
return 0;
}


This solution contains 42 empty lines, 64 comments and 6 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 29.7 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 207 MByte.

(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

December 10, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 35% (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.

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