The Trailing Number of Zeros of $n!$ is given by:

$\quad z(n) = \left\lfloor \space \dfrac {n}{5} \space \right\rfloor + \left\lfloor \space \dfrac {n}{5^2} \space \right\rfloor + \left\lfloor \space \dfrac {n}{5^3} \space \right\rfloor + \left\lfloor \space \dfrac {n}{5^4} \space \right\rfloor + ...$

where $\lfloor \space \rfloor$ is the greatest integer function.

For example, for $13!: \quad z(13) = \lfloor \frac {13}{5} \rfloor = 2$

This is because a trailing 0 is formed by a factor of $10=2\cdot 5,$ but there are always more 2's than 5's in the factorial, so we just need to count the number of powers of 5 in the product.

Similarly, for $1000!:$

$z(1000) = \lfloor \frac {1000}{5} \rfloor + \lfloor \frac {1000}{25} \rfloor + \lfloor \frac {1000}{125} \rfloor + \lfloor \frac {1000}{625} \rfloor = 200 + 40 + 8 + 1 = \boxed {249}$

A special case of de Polignac's (or Legendre's) Formula for trailing zeroes gives us:

$f(n)=\sum\limits_{i=1}^k \left \lfloor{n/5^i}\right \rfloor = \left \lfloor{n/5}\right \rfloor+\left \lfloor{n/5^2}\right \rfloor+\left \lfloor{n/5^3}\right \rfloor+\left \lfloor{n/5^4}\right \rfloor+...+\left \lfloor{n/5^k}\right \rfloor$ .

Plugging $1000$ in for $n$ , we have $200+40+8+1=249$ , our final answer.

Very simple solution:

To form a 0 you need a 2 and a 5 multiplied. It is obvious that there are more 5s than 2s, so the number of 0s is equal to the number of 5s.

• Numbers that have 1 five: 1000/5 = 200
• Number that have 2 fives: 1000/25 = 40
• Number that have 3 fives: 1000/125 = 8
• Number that have 4 fives: 1000/25 = 1

Number of 5s = Number of trailing 0s = 200 + 40 + 8 + 1 = $\boxed{249}$

(1000)! = 402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

No. of trailing zeroes = 249

The number of zeros can be written as a product of 2's and 5's Number of zeros = a ;where 'a' is (5*2)^a The exponent of 5 in 1000! is

1000/5 + 1000/25 + 1000/125 + 1000/625(rounds off to 1 )

= 200+40+8+1=249 Since 1000! Has more 2 exponents than 5 , the limiting number will be 5. Hence 1000! Will have 249 zeros Hope you found it easy:)

I have no idea but use a C++ program to solve it. The sourse code is following.

include<iostream>

define MAX 100000

using namespace std; int main() { int n,a[MAX]; int i,j,k,count,temp; while(cin>>n) { a[0]=1; count=1; for(i=1;i<=n;i++) { k=0; for(j=0;j<count;j++) { temp=a[j]*i+k; a[j]=temp%10; k=temp/10;
} while(k) { a[count++]=k%10; k/=10; } } for(j=MAX-1;j>=0;j--)
if(a[j]) break; for(i=count-1;i>=0;i--) cout<<a[i]; cout<<endl; } return 0; } Then we count these trailing zeroes and the problem can be solve.

More rudimentary form of Chew's solution. We know or observe that the trailing 0s are generated by multiples of 5 and only multiples of 5. Then: $1000!=X*5^{1000/5}*(1000/5)!=X*5^{200}*200!$ , where X is the product of all numbers that don't divide by 5 We continue this method for the subsequent factorials: $1000!=X*5^{200}*200! =X*Y*5^{200}*5^{40}*40!$ $1000!=X*5^{200}*200! =X*Y*5^{200}*5^{40}*40! =X*Y*Z*5^{200}*5^{40}*5^8*8!$ $1000!=X*5^{200}*200! =X*Y*5^{200}*5^{40}*40! =X*Y*Z*5^{200}*5^{40}*5^8*8!=X*Y*Z*W*5^{200}*5^{40}*5^8*5^1*1!$ Finally you add the power of 5 and you get the number of trailing zeros: $200+40+8+1 = 249$