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

# Problem 160: Factorial trailing digits

(see projecteuler.net/problem=160)

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.

That little trick means that f(1000000000000) = 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.

# My code

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

#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
unsigned long long 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.

# 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 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)*

# Benchmark

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

July 13, 2017 added comments

# Difficulty

Project Euler ranks this problem at **60%** (out of 100%).

# Links

projecteuler.net/thread=160 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

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My username at Project Euler is

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# 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 159 - Digital root sums of factorisations | Triominoes - problem 161 >> |