<< problem 539 - Odd elimination Robot Welders - problem 563 >>

# Problem 549: Divisibility of factorials

The smallest number m such that 10 divides m! is m=5.
The smallest number m such that 25 divides m! is m=10.

Let s(n) be the smallest number m such that n divides m!.
So s(10)=5 and s(25)=10.
Let S(n) be sum{s(i)} for 2 <= i <= n.
S(100)=2012.

Find S(10^8).

# My Algorithm

As always I wrote a simple function to solve the problem for small values.
The function naive(n) returns s(n) and easily verifies that s(2) + s(3) + s(4) + ... + s(100) = 2012.
Unfortunately it way too slow to find s(10^8) in a reasonable amount of time.
(If you had enough spare time it would produce the correct result eventually).

naive(n) computes 1! mod n, then 2! mod n, 3! mod n, ... until it finds a value result such that result! mod n == 0.

It took me a while to realize that
s(n) = s(p^{e_p} * other) = max(s(p^{e_p}), s(other) )

In plain English: if I factorize n into its prime factors p_1, p_2, ... then I only need to find a way to compute s(p^{e_p})
where e_p is the exponent of the prime factors.
An example:
24 = 2^3 * 3^1
s(24) = s(2^3 * 3) = max(s(2^3), s(3)) = max(4, 3) = 4

Resolving the recursive structure of the formula shown above:
s(n) = max(s(p_1^{e_1}), s(p_2^{e_2}), s(p_3^{e_3}), ...)
where p_i are the prime factors of n and e_i the exponents of those prime factors.

Obviously for each prime p we have s(p^1) = p because p! is the smallest factorial which contains p and thus can be divided by p.
A modified version of the naive algorithm is pretty fast when it comes to finding s(p^{e_p}):
instead of looking at each consecutive factorial 1!, 2!, 3!, ... I only look at each factorial that is a multiple of p:
p! mod p^{e_p}, (2p)! mod p^{e_p}, (3p)! mod p^{e_p}, ... until I find some (x * p)! mod p^{e_p} == 0.
My cache contains only 2633 such values prime^{2 ... x} < 10^8.

Whenever the main() function finds a prime number, then it adds its powers to the cache.
For all composite numbers it calls getSmallestFactorial which performs a prime factorization and returns the maximum value of any prime power encountered.

## Alternative Approaches

It's possible to write faster solutions using some special properties of the Kempner function (see en.wikipedia.org/wiki/Kempner_function).
However, I wasn't aware of it and therefore didn't look it up.

## Note

I'm surprised that this problem has a rating of only 10%. There are many easier problems with a higher percentage.
In my personal opinion its rating should be something like 40%.

Copying all prime numbers to a dense std::vector gives a little speed boost (almost 3x) in getSmallestFactorial at the cost of about 40 MByte RAM consumption.
The high execution time (about 30 seconds) combined with an increased memory usage puts my solution in the top spot of the "most expensive solutions" (as of August 2017) -
I didn't expect that when I saw the 10% rating ...

# 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 | ./549

Output:

Note: the original problem's input 100000000 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 <unordered_map>
#include <algorithm>

// ---------- 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) + 1;

// allocate memory
sieve.resize(half, true);
// 1 is not a prime number
sieve = 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 ----------

// compute all factorials until factorial % n == 0
unsigned int naive(unsigned int n)
{
unsigned long long factorial = 1;
unsigned int result = 0;
while (factorial % n != 0)
{
result++;
factorial *= result;
factorial %= n;
}
return result;
}

// all prime numbers < 10^8
std::vector<unsigned int> primes;
// cache for i^2, i^3, i^4, ... where i is prime
std::unordered_map<unsigned int, unsigned int> cache;

// compute s(n)
unsigned int getSmallestFactorial(unsigned int n)
{
// will be the result
unsigned int best = 0;

// split off all prime factors
for (auto p : primes)
{
// p is not a prime factor of the current number ?
if (n % p != 0)
continue;

// extract the current prime factor as often as possible
// e.g. => 24 => 2^3 * 3 => primePower will be 8 and reduced = 3
unsigned int primePower = 1;
do
{
n          /= p;
primePower *= p;
} while (n % p == 0);

// higher result ?
best = std::max(best, cache[primePower]);

// no further factorization possible ?
if (n == 1)
return best;
if (isPrime(n))
// s(prime) = prime
return std::max(best, n);
}

return best;
}

int main()
{
unsigned int limit = 100000000;
std::cin >> limit;

unsigned long long sum = 0;

// simple algorithm, too slow
//for (unsigned int i = 2; i <= 100; i++)
//  sum += naive(i);

// and now the more sophisticated approach

// find all primes below 10^8
fillSieve(limit);
// copy those 5761455 primes to a dense array for faster access
for (unsigned int i = 2; i < limit; i++)
if (isPrime(i))
primes.push_back(i);

// find result for numbers with are powers of a single prime
for (unsigned int i = 2; i <= limit; i++)
{
if (isPrime(i))
{
// pre-compute all values of i^2, i^3, ... where i is prime and store in cache[]
unsigned long long power = i * (unsigned long long) i;
for (unsigned int exponent = 2; power <= limit; exponent++)
{
// optimized version of naive(), skip i numbers in each iteration
unsigned long long factorial = i;
unsigned int result = i;
do
{
result    += i;
factorial *= result;
factorial %= power;
} while (factorial % power != 0);

cache[power] = result;

// next exponent
power *= i;
}

// s(prime) = prime
sum += i;
}
else
{
// compute s(non prime)
sum += getSmallestFactorial(i);
}
}

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


This solution contains 26 empty lines, 38 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 26.8 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 41 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

August 10, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 10% (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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