<< problem 159 - Digital root sums of factorisations Triominoes - problem 161 >>

# Problem 160: Factorial trailing digits

For any N, let f(N) be the last five digits before the trailing zeroes in N!.
For example,

9! = 362880 so f(9)=36288
10! = 3628800 so f(10)=36288
20! = 2432902008176640000 so f(20)=17664

Find f(1,000,000,000,000)

# My Algorithm

Wow, this was probably the most trial'n'error based algorithm so far.
I knew the correct answer (just entered "1000000000000!" in WolframAlpha) but my code was very slow.

It was obvious to me that a zero is appended to a factorial whenever the current number is a multiple of 5.
Actually, some of the previous numbers in the factorial needs to be a multiple of 2 because only 2*5=10 produces a zero,
but there are far more numbers which are a multiple of 2 than a multiple of 5.

The concept was to multiply all odd number (except those ending with 5) modulo 10000.
Then multiply all even numbers, but divided by 2.
Count the number of multiples of 2 (let's call them "twos") and count the multiple of 5 ("fives")
and multiply the result by 2^{twos-fives}.

My first attempts needed a few hours to display the correct result but using smaller factorials (which I verified with WolframAlpha, too)
I played around with the code which parts I can optimize in a true trial'n'error way:

• then I grouped ten consecutive numbers and saved a few modulo operations ...
• then I reduced all numbers that are neither a multiple of 2 nor 5 to their last five digits ...
• then I discovered that actually only the last digit of those numbers matters and in each iteration I can multiply by 1*3*7*9 right away
• then I saw that all multiples of 4 provides enough "twos" for the "fives", thus 1*2*3*6*7*8*9 is possible; only treat 4, 5, and 10 (=0) separately
• then I included the powmod function from my toolbox and did powmod(1*2*3*6*7*8*9, 10^12/10) once at the beginning
... and then my code algorithm1() needed just 12 minutes to find the correct result. Still too slow.

I knew that f(10^12) != f(10^11). More or less by chance I tried f(10^12 / 2) and f(10^12 / 5).
It turns out that f(10^12) = f(10^12 / 5) (but f(10^12) != f(10^12 / 2).
Even more, f(5N) = f(N) for 5N > Modulo. This can be applied repeatedly: f(5^k * N) = f(N) for 5N > Modulo.

My super-optimized code computes f(2560000) in less than 10 milliseconds. Yeah !

If I had known that trick right from the start then I would have probably only written algorithm2().
But I really enjoyed the ride of optimizing algorithm1() !

## Alternative Approaches

True mathematicians solved this problems with a weird series of modulo computations, Wilson's theorem (which I've never heard of) and many more.
It seems to me that problem 160 is among those problems with the biggest variety of solution strategies.

## Note

Most of my optimizations of algorithm1() focus on solving f(N) for large N.
It produces wrong results for small N.

# 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 20 | ./160

Output:

Note: the original problem's input 1000000000 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>

// last five digits
const unsigned int Modulo = 100000;

// ---------- powmod was taken from my toolbox ----------

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

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

// ---------- and now my first "brute force" algorithm ----------

unsigned int algorithm1(unsigned long long limit)
{
// process ten numbers in one iteration
unsigned long long iterations = limit / 10;

// count "excess"/"spare" twos that can be used whenever I encounter a five (2*5=10)
unsigned long long twos = 0;

// find last five digits of (2*3*6*7*8*9)^iterations
// note: the 4 is missing because sometimes I need a few "spare" twos
unsigned long long result = powmod(1*2*3*6*7*8*9, iterations, Modulo);

// process numbers 10i+1, 10i+2, 10i+3, ..., 10i+10
for (unsigned long long i = 0; i < iterations; i++)
{
// enough "spare" twos ?
if (twos > 100) // 100 was chosen "at random" but even 10 works fine
// yes, no need to hold back a few twos (2*2=4)
result *= 4;
else
// not enough "spare" twos, avoid multiplying by 4 and hold back 2 twos
twos   += 2;

// five = (i*10 + 5) / 5 = i*2 + 1
auto five = i*2 + 1;
// need one "spare" two
twos--;
// as long as it's still a multiple of 5 ...
while (five % 5 == 0)
{
five /= 5;
twos--; // and need one more "spare" two
}
result *= five % Modulo;

// ten = (i*10 + 10) = i + 1
auto ten = i + 1;
// note: unlike five, now no "spare" two is needed

// same loop as used for five ...
while (ten % 5 == 0)
{
ten /= 5;
twos--; //
}
result *= ten % Modulo;

// keep only the last five digits
result %= Modulo;
}

// multiply with remaining "spare" twos
while (twos-- > 0)
result = (result * 2) % Modulo;

return result;
}

// ---------- a very basic algorithm, sufficient for f(2560000) ----------

unsigned int algorithm2(unsigned long long limit)
{
unsigned long long result = 1;
// multiply all numbers
for (unsigned long long i = 1; i <= limit; i++)
{
auto current = i;
// divisible by 5 ?
while (current % 5 == 0)
{
current /= 5;
// there were at least the same number of twos
result  /= 2;
}

result *= current % Modulo;
// note: I can't use % Modulo because whenever a 5 appear, I divide by 2, too
//       and therefore can't erase too many digits higher than 10^5
result %= 1000000000ULL;
}

// reduce to five digits
return result % Modulo;
}

int main()
{
// 10^12
auto limit = 1000000000000ULL;
std::cin >> limit;

// f(5^k * n) = f(n) until n%10000 != 0
while (limit % Modulo == 0)
limit /= 5;

if (limit > 2560000)
// faster algorithm
std::cout << algorithm1(limit) << std::endl;
else
// simpler algorithm
std::cout << algorithm2(limit) << std::endl;

return 0;
}


This solution contains 23 empty lines, 34 comments and 1 preprocessor command.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.02 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 13, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 60% (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 159 - Digital root sums of factorisations Triominoes - problem 161 >>
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