# 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) + …Then answer is f(N)+f(M)

## C++

`// CPP program for count number of trailing zeros ` `// in N! * M! ` `#include <iostream> ` `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; ` `} ` |

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## 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.*/` |

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## 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. ` |

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## 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.*/` |

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## PHP

`<?php ` `// PHP program for count number ` `// of trailing zeros in N! * M! ` ` ` `// Returns number of ` `// zeros in factorial n ` `function` `trailingZero(` `$x` `) ` `{ ` ` ` ` ` `// dividing x by powers of ` ` ` `// 5 and update count ` ` ` `$i` `= 5; ` `$count` `= 0; ` ` ` `while` `(` `$x` `> ` `$i` `) ` ` ` `{ ` ` ` `$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. ` `?> ` |

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Output:

37

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