### algorithm

#### Finding natural numbers having n Trailing Zeroes in Factorial

I need help with the following problem. Gives an integer m and i need to find the number of positive integers n, such that the factorial of n ends with exactly m zeroes. I wrote this code it works fine and i get the right output, but it take way too much time as the numbers increase. a=input() while a: x=[] m,n,fact,c,j=input(),0,1,0,0 z=10*m t=10**m while z-1: fact=1 n=n+1 for i in range(1,n+1): fact=fact*i if fact%t==0 and ((fact/t)%10)!=0: x.append(int( n)) c=c+1 z=z-1 for p in range(c): print x[p], a-=1 print c Could someone suggest me a more efficient way to do this. Presently, it take 30 sec for a test case asking for numbers with 250 trailing zeros in its factorial. Thanks

To get number of trailing zeroes of n! efficiently you can put def zeroes(value): result = 0; d = 5; while (d <= value): result += value // d; # integer division d *= 5; return result; ... # 305: 1234! has exactly 305 trailing zeroes print zeroes(1234) In order to solve the problem (what numbers have n trailing zeroes in n!) you can use these facts: number of zeroes is a monotonous function: f(x + a) >= f(x) if a >= 0. if f(x) = y then x <= y * 5 (we count only 5 factors). if f(x) = y then x >= y * 4 (let me leave this for you to prove) Then implement binary search (on monotonous function). E.g. in case of 250 zeroes we have the initial range to test [4*250..5*250] == [1000..1250]. Binary search narrows the range down into [1005..1009]. 1005, 1006, 1007, 1008, 1009 are all numbers such that they have exactly 250 trainling zeroes in factorial

Focus on the number of 2s and 5s that makes up a number. e.g. 150 is made up of 2*3*5*5, there 1 pair of 2&5 so there's one trailing zero. Each time you increase the tested number, try figuring out how much 2 and 5s are in the number. From that, adding up previous results you can easily know how much zeros its factorial contains. For example, 15!=15*...*5*4*3*2*1, starting from 2: Number 2s 5s trailing zeros of factorial 2 1 0 0 3 1 0 0 4 2 0 0 5 2 1 1 6 3 1 1 ... 10 5 2 2 ... 15 7 3 3 .. 24 12 6 6 25 12 8 8 <- 25 counts for two 5-s: 25 == 5 * 5 == 5**2 26 13 8 8 .. Refer to Peter de Rivaz's and Dmitry Bychenko's comments, they have got some good advices.

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