Project Euler/20

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Project Euler Problem 20 is a problem related to factorial. Find the "digit sum" of the factorial of 100. To learn more about "digit sum", click here.

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Implementations

Aheui

The decimal multiplication algorithm has been improved and now it works for an arbitrarily large number multiplied by any number below a million in any integer base below a million.

밤발발따따빠박따뿌터너벌
ㅇㅇㅇ쑥너벌벌머차바싹볼
희ㅇㅇ뿌터너벌벌서
몽다석차바싼숙머오쎤석러더벌벌썩ㅇㅇ너
ㅇㅇㅇㅇ무선빠사빠싹삭따싼산다빠발발도
ㅇㅇ먀썩뻐ㅇㅇㅇ요

C

#include <stdio.h>
int a[320]={1},k,l,r;

void flow(int d[]){
    int i;
    for(i=0;i<319;++i){
        d[i+1]+=d[i]/10;
        d[i]%=10;
    }
}
int main(){
	for(k=1;k<=100;++k){
		for(l=0;l<320;++l){
            	 a[l]*=k;
        	}
        	flow(a);
	}
	for(l=0;l<320;++l){
		r+=a[l];
	}
	printf("%d",r);
    return 0;
}

I fuck, you fuck

There's a fucker named a
There's a fucker named b
Fuck b 100 times
while b is fucked
b fucks a again
I unfuck b
end fuck
There's a fucker named c
while a is fucked
c unfucks c
Fuck c 10 times
a unfucks c over and over
c fucks b
c unfucks c
Fuck c 10 times
a unfucks c again
a unfucks a
c fucks a
end fuck
b fucks you

Python

# Euler Problem 20
# by Europe2048

import math
def sumdigits(n):
	sum = 0
	for i in str(n):
		sum = sum + int(i)
	return sum
print(sumdigits(math.factorial(100)))

Alternate implenentation (shorter), by None1:

print(sum(map(int,str(__import__("math").factorial(100)))))

Version by Xi-816:

print(sum([int(i)for i in __import__("math").factorial(100)]))

External resources

• A004152, a related sequence on OEIS. The 101st term is the solution.
• Problem 20 on Project Euler Official Website (not available)
• Problem 20 on Project Euler Mirror