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# 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 : 8```

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 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

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 2a * a! * b!
D(n) = 2a * D(a) * D(b)
Implementation:

## C++

 `// C++ program to find last non-zero digit in n!``#include``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;``}`

## C

 `// C program to find last non-zero digit in n!``#include` `// 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;``    ``printf``(``"%d"``,lastNon0Digit(n));``    ``return` `0;``}` `// This code is contributed by allwink45.`

## 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

 ``

## Javascript

 ``

Output

`2`

Time complexity: O(log n)

Space complexity: O(log n)

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

Approach:-

1. 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.
2. Now what we can do is divide each array element into its shortest divisible form by 5 and increase count of such occurrences.
3. 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 of 5’s).
4. Multiply each number(after removing pairs of 2’s and 5’s) now and store just last digit by taking remainder by 10.
5. Now call recursively for smaller numbers by (currentNumber – 1) as parameter.

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `// Helper Function to return the rightmost non-zero digit``void` `callMeFactorialLastDigit(``int` `n, ``int` `result[], ``int` `sumOf5)``{``  ``int` `number = n; ``// assigning 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 = (result * (number % 10)) % 10;` `  ``// calling again for (currentNumber - 1)``  ``callMeFactorialLastDigit(n - 1, result, sumOf5);``}` `int` `lastNon0Digit(``int` `n)``{``  ``int` `result[] = { 1 }; ``// single element array.``  ``callMeFactorialLastDigit(n, result, 0);``  ``return` `result;``}` `int` `main()``{``  ``cout << lastNon0Digit(7) << endl;``  ``cout << lastNon0Digit(12) << endl;``  ``return` `0;``}` `// This code is contributed by rameshtravel07.`

## C

 `#include ` `// Helper Function to return the rightmost non-zero digit``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;``    ``sumOf5--;``}` `    ``/*multiplied result and current number(after``    ``removing pairs) and do modular division to get the``    ``last digit of the resultant number.*/``result = (result * (number % 10)) % 10;` `   ``// calling again for (currentNumber - 1)``callMeFactorialLastDigit(n - 1, result, sumOf5);``}` `int` `lastNon0Digit(``int` `n)``{``int` `result[] = { 1 }; ``// single element array``callMeFactorialLastDigit(n, result, 0);``return` `result;``}` `int` `main()``{``printf``(``"%d\n"``,lastNon0Digit(7));``printf``(``"%d"``,lastNon0Digit(12));``return` `0;``}`

## 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.`

## Python3

 `# Helper Function to return the rightmost non-zero digit``def` `callMeFactorialLastDigit(n, result, sumOf5):``    ``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 ``=` `int``(number ``/` `5``)``        ``# increase count of 5``        ``sumOf5 ``+``=` `1` `    ``# Divide the number by``    ``# 2 as much as possible``    ``while` `(sumOf5 !``=` `0` `and` `(number & ``1``) ``=``=` `0``):``        ``number >>``=` `1` `# dividing the number by 2``        ``sumOf5 ``-``=` `1` `    ``"""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)` `def` `lastNon0Digit(n):``    ``result ``=` `[ ``1` `] ``# single element array.``    ``callMeFactorialLastDigit(n, result, ``0``)``    ``return` `result[``0``]` `print``(lastNon0Digit(``7``)) ``# 3040``print``(lastNon0Digit(``12``)) ``# 479001600` `# This code is contributed by suresh07.`

## C#

 `using` `System;``class` `GFG {``    ` `    ``// Helper Function to return the rightmost non-zero``    ``// digit``    ``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 = (result * (number % 10)) % 10;``      ` `        ``// calling again for (currentNumber - 1)``        ``callMeFactorialLastDigit(n - 1, result, sumOf5);``    ``}`` ` `    ``static` `int` `lastNon0Digit(``int` `n)``    ``{``        ``int``[] result = { 1 }; ``// single element array.``        ``callMeFactorialLastDigit(n, result, 0);``        ``return` `result;``    ``}``    ` `  ``static` `void` `Main() {``    ``Console.WriteLine(lastNon0Digit(7)); ``// 3040``    ``Console.WriteLine(lastNon0Digit(12)); ``// 479001600``  ``}``}` `// This code is contributed by mukesh07.`

## Javascript

 ``

Output

```4
6```

Time complexity :- O(N)

Space complexity :- O(1)

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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