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Given an integer 1 ≤ N ≤ 1,000,000 as input, output the last non-zero digit of N!, where ! is the factorial (the product of all numbers from 1 to N, inclusive). This is OEIS sequence A008904.

Your program needs finish within 10 seconds on a reasonable machine for any valid input.

Test Cases

1 => 1
2 => 2
3 => 6
4 => 4
5 => 2
6 => 2
7 => 4
8 => 2
9 => 8
10 => 8
100 => 4
1000 => 2
10000 => 8
100000 => 6
1000000 => 4

This is a so the shortest code in bytes wins!

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  • \$\begingroup\$ Single function or complete program? \$\endgroup\$ – Joey Feb 8 '11 at 12:15
  • \$\begingroup\$ @joey No, they are just test cases. Single Input, Single Output. \$\endgroup\$ – fR0DDY Feb 8 '11 at 14:09
  • \$\begingroup\$ @joey Complete program. \$\endgroup\$ – fR0DDY Feb 8 '11 at 14:36
  • 1
    \$\begingroup\$ Requirement for a full program is discouraged... \$\endgroup\$ – Erik the Outgolfer Jan 18 '18 at 16:12
  • 2
    \$\begingroup\$ @EriktheOutgolfer this is from ~7 years ago, so I don't think that was determined at the time \$\endgroup\$ – NoOneIsHere Jan 18 '18 at 22:35

18 Answers 18

10
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Ruby - 63 chars

f=->n{n<2?1:6*[1,1,2,6,4,4,4,8,4,6][n%10]*3**(n/5%4)*f[n/5]%10}

Source - http://oeis.org/A008904

Handles f upto a thousand digits under a second.

Test

irb(main):014:0> for n in 2..6
irb(main):015:1> puts f[10**n]
irb(main):016:1> end
4
2
8
6
4
| improve this answer | |
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11
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Mathematica, 45 36 bytes

Last@Select[IntegerDigits[#!],#>0&]&

Very readable for a winning answer. :) (Then again, there's no GolfScript & Co. submission yet.)

This handles input 1,000,000 in about 5 seconds on my machine.

| improve this answer | |
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  • 1
    \$\begingroup\$ Mathematica is pretty much the perfect language for this question. \$\endgroup\$ – Michael Stern Feb 2 '15 at 0:03
4
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Python - 75

n=input()
g=1
while n:
 g*=n
 while g%10<1:g/=10
 g%=10**9
 n-=1
print g%10
| improve this answer | |
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3
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PARI/GP - 27 bytes

This trades speed for size -- the testcase takes a long time (~6 seconds).

n->n!/10^valuation(n!,5)%10

This version is much faster (~15 microseconds) but takes 81 bytes:

n->r=1;while(n,r*=Mod(4,10)^(n\10%2)*[1,2,6,4,2,2,4,2,8][max(n%10,1)];n\=5);lift(r)

You can use this (non-golfed) code to test either:

[%(10^n) | n <- [1..6]]
| improve this answer | |
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2
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Windows PowerShell, 53 56 59 60 63 73 90

($a=1).."$input"|%{$a="$($a*$_)".trim('0')%1e7}
$a%10

Notes:

  • Takes longer than a minute for a number close to 100,000. However, removing zeroes at the end requires a convert to string, doing calculations requires a number, so the conversions are inevitable in any case.

History:

  • 2011-02-08 10:31 (90) – First attempt.
  • 2011-02-08 10:33 (73) – Modulus is shorter than slicing and joining.
  • 2011-02-08 10:34 (63) – Unnecessary trim.
  • 2011-02-08 10:37 (60) – Unnecessary cast to a number. Modulus does that just fine, already.
  • 2011-02-08 10:40 (59) – Some inlining.
  • 2011-02-08 11:00 (56) – What did I say before about modulus being shorter? Applies to the output as well.
  • 2011-02-08 11:01 (53) – Casting $input to a string is enough; the cast to int is applied implicitly.
| improve this answer | |
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2
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Perl, 53 58 61 characters

All whitespace can be removed, but I left it in for "readability". Note: not using some silly explicit formula from Sloane.

sub f {
    $_ = $1 * ++$n || 1, /(.{1,7}?)0*$/ while $n < $_[0];
    $1 % 10
}

Calculates f(10^6) in 8.7 seconds on my machine.

Update: OP wanted it to be a whole program:

$_ = $1 * ++$n || 1, /(.{1,7}?)0*$/ while $n < $ARGV[0];
print $1 % 10

That makes it 55 characters.

| improve this answer | |
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2
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CJam - 28

1ri{I)*_AbW%{}#A\#/1e7%}fIA%

You can try it at http://cjam.aditsu.net/ for values up to 10000 or so; for larger numbers you should use the java interpreter. 1000000 runs in about 3 seconds on my laptop.

Explanation:

Unfortunately the straightforward solution is too slow, so I'm keeping only the last 7 digits (before the trailing zeros) after each multiplication.

1           push 1 on the stack
ri          read a token and convert to integer
{           loop (for I from 0 to N - 1)
    I)      push I and increment
    *       multiply with the previous value (initially 1)
    _Ab     duplicate and convert to array of digits
    W%      reverse array
    {}#     find the position of the first non-zero digit
    A\#     raise 10 to that power
    /       divide, thus removing all trailing zeros
    1e7%    keep the remainder modulo 10000000
}fI         end for loop
A%          get the last digit

Note: this language is a lot newer than the question.

| improve this answer | |
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2
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Mathematica, 34 bytes

Mod[#!/10^IntegerExponent[#!],10]&
| improve this answer | |
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2
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Java (OpenJDK 8), 62 bytes

n->{long f=n;for(;n>1||f%10==0;)f=n>1?f*--n:f/10;return f%10;}

Try it online!

Similar to @Kevin Cruijssen but saves 5 bytes by combining the loops.

| improve this answer | |
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  • \$\begingroup\$ Welcome to PPCG! Nice first post! Hope you stick around! \$\endgroup\$ – Rɪᴋᴇʀ Jan 19 '18 at 1:11
  • \$\begingroup\$ Welcome to PPCG! I agree with @Riker, great first post. Well done golfing my code by combining the loops. You can golf 1 more byte in your current answer by replacing || with |, and an additional byte by replacing ==0 with <1. Enjoy your stay! \$\endgroup\$ – Kevin Cruijssen Feb 1 '18 at 12:19
2
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C, 150 140 135 bytes

r,d;f(k,x){r=x<5?3:f(k+1,x/5);return(d=x%5)?r*"33436"[d]*(1<<d*k%4)%5:r;}main(int c,char**v){c=atoi(*++v);printf("%d",c<2?1:2*f(0,c));}

This is the version for ASCII systems; replace the string 33436 with 11214 for an EBCDIC system, or with \1\1\2\1\4 for a portable program.

C solutions are a bit hampered by the requirement to provide a full program; however, this does answer the question fully.

Try it online (requires Javascript):

Explanation

It's based on the algorithm outlined in Least Significant Non-Zero Digit of n!, turned around so that we recurse in to find the highest power of five, and do the calculation on the way out. The tables of constants were too big, so I reduced them by finding a relationship between the previous residue r, the current digit d and the recursion depth k:

     0    1       2       3    4  =d
  0  0  3×2^k  1×2^2k  3×2^3k  2
  1  1  1×2^k  2×2^2k  1×2^3k  4
r 2  2  2×2^k  4×2^2k  2×2^3k  3
  3  3  3×2^k  3×2^2k  3×2^3k  2
  4  4  4×2^k  4×2^2k  4×2^3k  1

For r>0, this resolves to a constant times r times 2^dk (mod 5); the constants are in a[] below (inlined in the golfed code). We also observe that (2^4)%5 is 1, so we can reduce the exponent to avoid overflowing the range of int.

const int a[] = { 1, 1, 2, 1, 4 };
int f(int k, int x){
    int r = x<5 ? 3 : f(k+1,x/5); /* residue - from recursing to higher-order quinary digits */
    int d = x%5;
    if (!d)
        return r;
    return r * a[d] * (1<<d*k%4) % 5;
}

int main(int c, char **v)
{
    c = atoi(*++v);
    printf("%d",
           c<2
           ? 1                  /* special-case 0 & 1 */
           : 2*f(0,c));         /* otherwise, it's 2 times r */
}

Tests:

$ for i in 100 1000 10000 100000; do echo $i: `./694 $i`; done
100: 4
1000: 2
10000: 8
100000: 6
1000000: 4

Performance is respectable, too. Here's a maximum input for a system with 32-bit int:

$ time ./694 2147483647
8
real    0m0.001s
user    0m0.000s
sys     0m0.000s

I got the same timings with a maximal 64-bit int, too.

| improve this answer | |
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  • 1
    \$\begingroup\$ It may be of interest to note that 2147483647! has over 19 billion digits, and (2^63-1)! over 170,000,000,000,000,000,000 digits, so this is a big win over calculating the factorials. 1000000! as specified in the question is feasible to calculate on current hardware; that's only 5½ million digits. :-) \$\endgroup\$ – Toby Speight May 17 '16 at 9:49
1
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PHP - 105

 <?foreach(explode("\n",`cat`)as$n)if($n){$f=rtrim(gmp_strval(gmp_fact($n)),'0');echo substr($f,-1)."\n";}

Runs under 10 seconds with the given testcase.

| improve this answer | |
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1
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Python3

239 Chars

Can do 10000 in ~3.2 seconds (Ideone cuts me off at 8 seconds, i'm sure it'll take longer than 10secs though :( )

from functools import *
N=100
r=range
s=(p for p in r(2,N)if all(p%n>0for n in r(2,p)))
f=lambda n,x:n//x+(n//x>0and f(n//x,x)or 0)
e=list([p,f(N,p)]for p in s)
e[0][1]-=e[2][1]
e[2][1]=0
print(reduce(lambda x,y:x*y,map(lambda x:x[0]**x[1],e))%10)

Python2.6

299 Chars (a bit faster)

from itertools import *
N=100000
r=xrange
def s(c=count(2)):
        while 1:p=c.next();c=ifilter(p.__rmod__,c);yield p
f=lambda n,x:n//x+(n//x>0and f(n//x,x)or 0)
e=[[p,f(N,p)]for p in takewhile(lambda x:x<N,s())]
e[0][1]-=e[2][1]
e[2][1]=0
print(reduce(lambda x,y:x*y,map(lambda x:pow(x[0],x[1],10),e))%10)
| improve this answer | |
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1
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Haskell, 78 characters

f n=head$dropWhile(=='0')$reverse$show$product[1..n]
main=interact(show.f.read)

(Would probably need to be compiled to compute 1,000,000! in 10 secs).

| improve this answer | |
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  • \$\begingroup\$ Save two chars, replace that foldl1 with product (cf codegolf.stackexchange.com/questions/607/find-the-factorial/…). But have you actually tried with 1000000! ? \$\endgroup\$ – J B Feb 8 '11 at 14:48
  • \$\begingroup\$ PS: not a complete program. \$\endgroup\$ – J B Feb 8 '11 at 14:49
  • \$\begingroup\$ Sorry, did it before that was clarified in the comments. I'll update it. \$\endgroup\$ – stusmith Feb 8 '11 at 16:09
  • \$\begingroup\$ main=do x<-readLn;print$head$snd$span(=='0')$reverse$show$product[1..x] saves 7 characters \$\endgroup\$ – Benji Oct 14 at 16:00
1
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J – 42 40 characters

A whole program. Save this program in a file and run with jconsole script.ijs 1234. Notice that this program does not exit the interpreter after printing a result. Type ^D or exit]0 to exit the interpreter.

echo([:{:@(#~*)10&#.inv@*)/1+i.".>{:ARGV

Here is an explanation:

  • x #. y interprets the integer vector y as a base x number; for example, 10 #. 1 2 3 4 yields 1234.
  • u inv yields the inverse of a verb u. In particular, x #. inv y represents y as a base x number; for example, 10 #. 1234 yields 1 2 3 4. Notice that inv is defined as ^:_1, that is, u applied -1 time.
  • x * y is the product of x and y, thus x 10&#.inv@* y yields a base-10 representation of the product of x and y.
  • x # y copies the n-th item of y as often as the n-th item of x; when x is a vector of booleans, x selects which items of y to take. For instance, 1 0 1 0 # 1 2 3 4 yields 1 3.
  • * y yields the signum of y.
  • x u~ y is the reflexive of u, that is, the same as y u x.
  • Thus, y #~ * y yields a vector of all items of y that are positive. In tacit notation, this can written with a hook as (#~ *).
  • {: y yields the last item in y.
  • assembled together, we get the tacit phrase ([:{:@(#~*)10&#.inv@*).
  • u/ y is the reduction of y, that is, the dyadic verb u inserted between elements of y. For instance, +/1 2 3 4 is like 1 + 2 + 3 + 4 and yields 10.
  • Thus, the phrase ([:{:@(#~*)10&#.inv@*)/ y yields the last digit of the product of the items of y.
  • ARGV is a boxed vector of the command line arguments.
  • ".>{:ARGV is the last argument unboxed and interpreted as a number.
  • i. y computes natural numbers from 0 to y - 1.
  • Thus, 1+i. y yields natural numbers from 1 to y. I could have also used >: increment here, but 1+ is clearer at the same cost of characters.
  • The entire program just applies 1+i.".>{:ARGV (the vector of 1 to the number in the last command line argument) to the verb ([:{:@(#~*)10&#.inv@*)/ and prints the result with echo.
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1
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Perl 6,  26  35 bytes

{[*](1..$_)~~/.*<(<-[0]>/}

Try it


As a full program:

put [*](1..@*ARGS[0])~~/.*<(<-[0]>/

Try it

Expanded:

{
  [*]( 1..$_ ) # reduce using &infix:« * »
  ~~           # match with
  /
    .*         # any number of values (so it matches from the end)
    <(         # only capture the following
    <-[0]>     # any value but 0 (negated character class)
  /
}
| improve this answer | |
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1
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C (gcc), 72 bytes (function)

f(n,d)long long n,d;{for(d=1;n;d%=10000)for(d*=n--;d%10<1;d/=10);d%=10;}

Try it online!

C (gcc), 101 99 bytes (whole program)

main(){long long n,d=1;for(scanf("%lld",&n);n;d%=10000)for(d*=n--;d%10<1;d/=10);printf("%d",d%10);}

Try it online!

This question is just shy of 8 years old so "reasonable machine" is not the same as back then, but I'm getting times of ~.01 seconds on my computer when doing all test cases together, so unless computers have increased in speed by a factor of 1000 this last decade, it should be fine.

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  • \$\begingroup\$ Moore's law is still (kinda) holding up, so it should be about x16 times faster \$\endgroup\$ – ASCII-only Jan 11 '19 at 0:51
  • \$\begingroup\$ Also, a function is fine \$\endgroup\$ – ASCII-only Jan 11 '19 at 0:51
1
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Attache, 64 bytes

{k.=Floor[_/5]If[_<2,1,6*Digits@$`U1sTny@(_%10)*3^(k%4)*$@k%10]}

Try it online!

Simple implementation of the OEIS formula. Digits@$`U1sTny compresses the hardcoded dictionary using Attache's compressed number feature. $ refers to the function itself within this answer.

| improve this answer | |
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  • \$\begingroup\$ This seems incredibly slow - the TIO timed out (1 minute) on the 100000 test case - how did you get the 1000000 result within 10 seconds? \$\endgroup\$ – Toby Speight Feb 20 '18 at 14:35
  • \$\begingroup\$ @TobySpeight When I answered this challenge, that particular requirement was not there (check the revision history). \$\endgroup\$ – Conor O'Brien Feb 20 '18 at 14:42
  • \$\begingroup\$ Ah, I should have looked at the history! You did verify all the testcases in the question, though? \$\endgroup\$ – Toby Speight Feb 20 '18 at 15:01
  • \$\begingroup\$ It seems there was a flurry of answers during the period that the time limit was removed from the question - that's unfortunate, really. \$\endgroup\$ – Toby Speight Feb 20 '18 at 15:09
  • \$\begingroup\$ @TobySpeight Yes, I did. It is unfortunate, and I'm not sure of the policy regarding this. \$\endgroup\$ – Conor O'Brien Feb 20 '18 at 16:07
0
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AWK, 47 57 bytes

{for(p=$1;--$1;p=(p*$1)%1e4)while(!(p%10))p/=10;$0=p%10}1

Try it online!

Original solution did not handle "large" input values very well. Could add -M to force it to work, but that also requires a lot more processing time.

| improve this answer | |
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  • \$\begingroup\$ Yeah, @TobySpeight , inf doesn't % very well. :( \$\endgroup\$ – Robert Benson Feb 20 '18 at 16:05
  • \$\begingroup\$ Ah... looking at the version of the question I answered, large numbers weren't required. \$\endgroup\$ – Robert Benson Feb 20 '18 at 16:14

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