# Trailing number of 0s in product of two factorials

Given two integer N or M find the number of zero’s trailing in product of factorials (N!*M!)?
Examples:

```Input : N = 4, M = 5
Output :  1
Explanation : 4! = 24, 5! = 120
Product has only 1 trailing 0.

Input : N = 127!, M = 57!
Output : 44```

As discussed in number of zeros in N! can be calculated by recursively dividing N by 5 and adding up the quotients.

For example if N = 127, then
Number of 0 in 127! = 127/5 + 127/25 + 127/125 + 127/625
= 25 + 5 + 1 + 0
= 31

Number of 0s in N! = 31. Similarly, for M we can calculate and add both of them.
So, by above we can conclude that number of zeroes in N!*M! Is equal to sum of number of zeroes in N! and M!.

f(N) = floor(N/5) + floor(N/5^2) + … floor(N/5^3) + …
f(M) = floor(x/5) + floor(M/5^2) + … floor(M/5^3) + …

## C++

 `// CPP program for count number of trailing zeros` `// in N! * M!` `#include ` `using` `namespace` `std;`   `// Returns number of zeros in factorial n` `int` `trailingZero(``int` `x)` `{` `    ``// dividing x by powers of 5 and update count` `    ``int` `i = 5, count = 0;` `    ``while` `(x > i) {` `        ``count = count + x / i;` `        ``i = i * 5;` `    ``}` `    ``return` `count;` `}`   `// Returns count of trailing zeros in M! x N!` `int` `countProductTrailing(``int` `M, ``int` `N)` `{` `   ``return` `trailingZero(N) + trailingZero(M);` `}`   `// Driver program` `int` `main()` `{` `    ``int` `N = 67, M = 98;` `    ``cout << countProductTrailing(N, M);` `    ``return` `0;` `}`

## Java

 `// Java program for count number` `// of trailing zeros in N! * M!` `import` `java.io.*;`   `class` `GFG {` `    `  `    ``// Returns number of zeros in factorial n` `    ``static` `int` `trailingZero(``int` `x)` `    ``{` `        ``// dividing x by powers ` `        ``// of 5 and update count` `        ``int` `i = ``5``, count = ``0``;` `        `  `        ``while` `(x > i) {` `            `  `            ``count = count + x / i;` `            ``i = i * ``5``;` `        ``}` `        ``return` `count;` `    ``}` `    `  `    ``// Returns count of trailing zeros in M! x N!` `    ``static` `int` `countProductTrailing(``int` `M, ``int` `N)` `    ``{` `    ``return` `trailingZero(N) + trailingZero(M);` `    ``}` `    `  `    ``// Driver program` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `N = ``67``, M = ``98``;` `        ``System.out.println(countProductTrailing(N, M));` `    ``}` `}`     `/* This code is contributed by Nikita Tiwari.*/`

## Python3

 `# Python 3 program for count` `# number of trailing zeros` `# in N ! * M !`   `# Returns number of zeros in` `# factorial n` `def` `trailingZero(x) :`   `    ``# Dividing x by powers of 5` `    ``# and update count` `    ``i ``=` `5` `    ``count ``=` `0` `    `  `    ``while` `(x > i) :` `        ``count ``=` `count ``+` `x ``/``/` `i` `        ``i ``=` `i ``*` `5` `    `  `    ``return` `count` `    `  `# Returns count of trailing ` `# zeros in M ! x N !` `def` `countProductTrailing(M, N) :` `    ``return` `trailingZero(N) ``+` `trailingZero(M)` `    `  `# Driver program` `N ``=` `67` `M ``=` `98` `print``(countProductTrailing(N, M))`   `# This code is contributed by Nikita Tiwari.`

## C#

 `// Java program for count number` `// of trailing zeros in N! * M!` `using` `System;`   `class` `GFG {` `    `  `    ``// Returns number of zeros in factorial n` `    ``static` `int` `trailingZero(``int` `x)` `    ``{` `        ``// dividing x by powers ` `        ``// of 5 and update count` `        ``int` `i = 5, count = 0;` `        `  `        ``while` `(x > i) {` `            `  `            ``count = count + x / i;` `            ``i = i * 5;` `        ``}` `        ``return` `count;` `    ``}` `    `  `    ``// Returns count of trailing zeros in M! x N!` `    ``static` `int` `countProductTrailing(``int` `M, ``int` `N)` `    ``{` `    ``return` `trailingZero(N) + trailingZero(M);` `    ``}` `    `  `    ``// Driver program` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `N = 67, M = 98;` `        ``Console.WriteLine(countProductTrailing(N, M));` `    ``}` `}`     `/* This code is contributed by vt_m.*/`

## PHP

 ` ``\$i``) ` `    ``{` `        ``\$count` `= ``\$count` `+ (int)(``\$x` `/ ``\$i``);` `        ``\$i` `= ``\$i` `* 5;` `    ``}` `    ``return` `\$count``;` `}`   `// Returns count of trailing ` `// zeros in M! x N!` `function` `countProductTrailing(``\$M``, ``\$N``)` `{` `    ``return` `trailingZero(``\$N``) + trailingZero(``\$M``);` `}`   `// Driver Code` `\$N` `= 67; ``\$M` `= 98;` `echo``(countProductTrailing(``\$N``, ``\$M``));`   `// This code is contributed by Ajit.` `?>`

## Javascript

 ``

Output:

` 37`

Time Complexity: O(log5m+log5n)

Auxiliary Space: O(1)

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