# Last non-zero digit of a factorial

Given a number n, find the last non-zero digit in n!.

Examples:

Input : n = 5 Output : 2 5! = 5 * 4 * 3 * 2 * 1 = 120 Last non-zero digit in 120 is 2. Input : n = 33 Output : 4

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A **Simple Solution** is to first find n!, then find the last non-zero digit of n. This solution doesn’t work for even slightly large numbers due to arithmetic overflow.

A **Better Solution** is based on the below recursive formula

Let D(n) be the last non-zero digit in n! If tens digit (or second last digit) of n is odd D(n) = 4 * D(floor(n/5)) * D(Unit digit of n) If tens digit (or second last digit) of n is even D(n) = 6 * D(floor(n/5)) * D(Unit digit of n)

**Illustration of **the **formula:**

For the numbers less than 10 we can easily find the last non-zero digit by the above simple solution, i.e., first computing n!, then finding the last digit.

D(1) = 1, D(2) = 2, D(3) = 6, D(4) = 4, D(5) = 2,

D(6) = 2, D(7) = 4, D(8) = 2, D(9) = 8.

D(1) to D(9) are assumed to be precomputed. Example 1: n = 27 [Second last digit is even]: D(27) = 6 * D(floor(27/5)) * D(7) = 6 * D(5) * D(7) = 6 * 2 * 4 = 48 Last non-zero digit is 8 Example 2: n = 33 [Second last digit is odd]: D(33) = 4 * D(floor(33/5)) * D(3) = 4 * D(6) * 6 = 4 * 2 * 6 = 48 Last non-zero digit is 8

**How does **the **above formula work?**

The below explanation provides intuition behind the formula. Readers may refer Refer http://math.stackexchange.com/questions/130352/last-non-zero-digit-of-a-factorial for complete proof.

14! = 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 Since we are asked about last non-zero digit, we remove all 5's and equal number of 2's from factors of 14!. We get following: 14! = 14 * 13 * 12 * 11 * 2 * 9 * 8 * 7 * 6 * 3 * 2 * 1 Now we can get last non-zero digit by multiplying last digits of above factors!

In n! a number of 2’s are always more than a number of 5’s. To remove trailing 0’s, we remove 5’s and equal number of 2’s.

Let **a **= floor(n/5), **b** = n % 5. After removing an equal number of 5’s and 2’s, we can reduce the problem from n! to 2^{a} * a! * b!

D(n) = 2^{a} * D(a) * D(b)**Implementation:**

## C++

`// C++ program to find last non-zero digit in n!` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Initialize values of last non-zero digit of` `// numbers from 0 to 9` `int` `dig[] = {1, 1, 2, 6, 4, 2, 2, 4, 2, 8};` `int` `lastNon0Digit(` `int` `n)` `{` ` ` `if` `(n < 10)` ` ` `return` `dig[n];` ` ` `// Check whether tens (or second last) digit` ` ` `// is odd or even` ` ` `// If n = 375, So n/10 = 37 and (n/10)%10 = 7` ` ` `// Applying formula for even and odd cases.` ` ` `if` `(((n/10)%10)%2 == 0)` ` ` `return` `(6*lastNon0Digit(n/5)*dig[n%10]) % 10;` ` ` `else` ` ` `return` `(4*lastNon0Digit(n/5)*dig[n%10]) % 10;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 14;` ` ` `cout << lastNon0Digit(n);` ` ` `return` `0;` `}` |

## Java

`// Java program to find last` `// non-zero digit in n!` `class` `GFG` `{` ` ` `// Initialize values of last non-zero digit of` ` ` `// numbers from 0 to 9` ` ` `static` `int` `dig[] = {` `1` `, ` `1` `, ` `2` `, ` `6` `, ` `4` `, ` `2` `, ` `2` `, ` `4` `, ` `2` `, ` `8` `};` ` ` ` ` `static` `int` `lastNon0Digit(` `int` `n)` ` ` `{` ` ` `if` `(n < ` `10` `)` ` ` `return` `dig[n];` ` ` ` ` `// Check whether tens (or second last)` ` ` `// digit is odd or even` ` ` `// If n = 375, So n/10 = 37 and` ` ` `// (n/10)%10 = 7 Applying formula for` ` ` `// even and odd cases.` ` ` `if` `(((n / ` `10` `) % ` `10` `) % ` `2` `== ` `0` `)` ` ` `return` `(` `6` `* lastNon0Digit(n / ` `5` `)` ` ` `* dig[n % ` `10` `]) % ` `10` `;` ` ` `else` ` ` `return` `(` `4` `* lastNon0Digit(n / ` `5` `)` ` ` `* dig[n % ` `10` `]) % ` `10` `;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `n = ` `14` `;` ` ` `System.out.print(lastNon0Digit(n));` ` ` `}` `}` `// This code is contributed by Anant Agarwal.` |

## Python3

`# Python program to find` `# last non-zero digit in n!` `# Initialize values of` `# last non-zero digit of` `# numbers from 0 to 9` `dig` `=` `[` `1` `, ` `1` `, ` `2` `, ` `6` `, ` `4` `, ` `2` `, ` `2` `, ` `4` `, ` `2` `, ` `8` `]` ` ` `def` `lastNon0Digit(n):` ` ` `if` `(n < ` `10` `):` ` ` `return` `dig[n]` ` ` ` ` `# Check whether tens (or second last) digit` ` ` `# is odd or even` ` ` `# If n = 375, So n/10 = 37 and (n/10)%10 = 7` ` ` `# Applying formula for even and odd cases.` ` ` `if` `(((n` `/` `/` `10` `)` `%` `10` `)` `%` `2` `=` `=` `0` `):` ` ` `return` `(` `6` `*` `lastNon0Digit(n` `/` `/` `5` `)` `*` `dig[n` `%` `10` `]) ` `%` `10` ` ` `else` `:` ` ` `return` `(` `4` `*` `lastNon0Digit(n` `/` `/` `5` `)` `*` `dig[n` `%` `10` `]) ` `%` `10` ` ` `return` `0` `# driver code` `n ` `=` `14` `print` `(lastNon0Digit(n))` `# This code is contributed` `# by Anant Agarwal.` |

## C#

`// C# program to find last` `// non-zero digit in n!` `using` `System;` `class` `GFG {` ` ` ` ` `// Initialize values of last non-zero` ` ` `// digit of numbers from 0 to 9` ` ` `static` `int` `[]dig = {1, 1, 2, 6, 4, 2, 2, 4, 2, 8};` ` ` ` ` `static` `int` `lastNon0Digit(` `int` `n)` ` ` `{` ` ` `if` `(n < 10)` ` ` `return` `dig[n];` ` ` ` ` `// Check whether tens (or second` ` ` `// last) digit is odd or even` ` ` `// If n = 375, So n/10 = 37 and` ` ` `// (n/10)%10 = 7 Applying formula` ` ` `// for even and odd cases.` ` ` `if` `(((n / 10) % 10) % 2 == 0)` ` ` `return` `(6 * lastNon0Digit(n / 5) *` ` ` `dig[n % 10]) % 10;` ` ` `else` ` ` `return` `(4 * lastNon0Digit(n / 5) *` ` ` `dig[n % 10]) % 10;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `int` `n = 14;` ` ` `Console.Write(lastNon0Digit(n));` ` ` `}` `}` `// This code is contributed by Nitin Mittal.` |

## PHP

`<?php` `// PHP program to find last` `// non-zero digit in n!` `// Initialize values of` `// last non-zero digit of` `// numbers from 0 to 9` `$dig` `= ` `array` `(1, 1, 2, 6, 4,` ` ` `2, 2, 4, 2, 8);` `function` `lastNon0Digit(` `$n` `)` `{` ` ` ` ` `global` `$dig` `;` ` ` `if` `(` `$n` `< 10)` ` ` `return` `$dig` `[` `$n` `];` ` ` `// Check whether tens(or second ` ` ` `// last) digit is odd or even` ` ` `// If n = 375, So n/10 = 37 and` ` ` `// (n/10)%10 = 7` ` ` `// Applying formula for even` ` ` `// and odd cases.` ` ` `if` `(((` `$n` `/ 10) % 10) % 2 == 0)` ` ` `return` `(6 * lastNon0Digit(` `$n` `/ 5) *` ` ` `$dig` `[` `$n` `% 10]) % 10;` ` ` `else` ` ` `return` `(4 * lastNon0Digit(` `$n` `/ 5) *` ` ` `$dig` `[` `$n` `% 10]) % 10;` `}` `// Driver code` `$n` `= 14;` `echo` `(lastNon0Digit(` `$n` `));` `// This code is contributed by Ajit.` `?>` |

## Javascript

`<script>` ` ` `// Javascript program to find` ` ` `// last non-zero digit in n!` ` ` ` ` `// Initialize values of last non-zero` ` ` `// digit of numbers from 0 to 9` ` ` `let dig = [1, 1, 2, 6, 4, 2, 2, 4, 2, 8];` ` ` ` ` `function` `lastNon0Digit(n)` ` ` `{` ` ` `if` `(n < 10)` ` ` `return` `dig[n];` ` ` ` ` `// Check whether tens (or second` ` ` `// last) digit is odd or even` ` ` `// If n = 375, So n/10 = 37 and` ` ` `// (n/10)%10 = 7 Applying formula` ` ` `// for even and odd cases.` ` ` `if` `((parseInt(n / 10, 10) % 10) % 2 == 0)` ` ` `return` `(6 * lastNon0Digit(parseInt(n / 5, 10))` ` ` `* dig[n % 10]) % 10;` ` ` `else` ` ` `return` `(4 * lastNon0Digit(parseInt(n / 5, 10))` ` ` `* dig[n % 10]) % 10;` ` ` `}` ` ` ` ` `let n = 14;` ` ` `document.write(lastNon0Digit(n));` ` ` `</script>` |

**Output**

2

** A Simple Solution based on recursion having worst-case Time Complexity O(nLog(n)).**

Approach:-

- It is given that you have to find the last positive digit. Now a digit is made multiple of 10 if there are 2 and 5. They produce a number with last digit 0.
- Now what we can do is divide each array element into its shortest divisible form by 5 and increase count of such occurrences.
- Now divide each array element into its shortest divisible form by 2 and decrease count of such occurrences. This way we are not considering the multiplication of 2 and a 5 in our multiplication(number of 2’s present in multiplication result upto n is always more than number 0f 5’s).
- Multiply each number(after removing pairs of 2’s and 5’s) now and store just last digit by taking remainder by 10.
- Now call recursively for smaller numbers by (currentNumber – 1) as parameter.

**Below is the implementation of the above approach: **

## Java

`/*package whatever //do not write package name here */` `import` `java.io.*;` `class` `GFG {` ` ` `// Helper Function to return the rightmost non-zero` ` ` `// digit` ` ` `public` `static` `void` ` ` `callMeFactorialLastDigit(` `int` `n, ` `int` `[] result,` ` ` `int` `sumOf5)` ` ` `{` ` ` `int` `number = n; ` `// assaigning to new variable.` ` ` `if` `(number == ` `1` `)` ` ` `return` `; ` `// base case` ` ` `// To store the count of times 5 can` ` ` `// divide the number.` ` ` `while` `(number % ` `5` `== ` `0` `) {` ` ` `number /= ` `5` `;` ` ` `// increase count of 5` ` ` `sumOf5++;` ` ` `}` ` ` `// Divide the number by` ` ` `// 2 as much as possible` ` ` `while` `(sumOf5 != ` `0` `&& (number & ` `1` `) == ` `0` `) {` ` ` `number >>= ` `1` `; ` `// dividing the number by 2` ` ` `sumOf5--;` ` ` `}` ` ` `/*multiplied result and current number(after` ` ` `removing pairs) and do modular division to get the` ` ` `last digit of the resultant number.*/` ` ` `result[` `0` `] = (result[` `0` `] * (number % ` `10` `)) % ` `10` `;` ` ` `// calling again for (currentNumber - 1)` ` ` `callMeFactorialLastDigit(n - ` `1` `, result, sumOf5);` ` ` `}` ` ` `public` `static` `int` `lastNon0Digit(` `int` `n)` ` ` `{` ` ` `int` `[] result = { ` `1` `}; ` `// single element array.` ` ` `callMeFactorialLastDigit(n, result, ` `0` `);` ` ` `return` `result[` `0` `];` ` ` `}` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `System.out.println(lastNon0Digit(` `7` `)); ` `// 3040` ` ` `System.out.println(lastNon0Digit(` `12` `)); ` `// 479001600` ` ` `}` `}` `//This code is contributed by KaaL-EL.` |

**Output**

4 6

we used single element array (int[] result = {1}) instead of integer as Java is Strictly Pass by Value!. It does not allow pass by reference for primitive data types. That’s why I used a single element array so that the recursive function can change the value of variable(result here). If we would have taken (int result = 1) then this variable remain unaffected.

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