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# Find the unit place digit of sum of N factorials

• Last Updated : 24 Mar, 2023

Given a number N, the task is to find units place digit of the first N natural numbers factorials, i.e. 1!+2!+3!+….N! where N<=10e18.
Examples:

```Input: n = 2
Output: 3
1! + 2! = 3
Last digit is 3

Input: n = 3
Output: 9
1! + 2! + 3! = 9
Last digit is 9```

Brute Force Approach:

In this approach, we are calculating the factorial of each number and then adding it to the sum. To prevent overflow, we are taking the modulo with 10 after every multiplication and addition operation. Finally, we return the unit place digit of the sum.

Below is implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std;`   `int` `get_unit_digit(``long` `long` `int` `N) {` `    ``if` `(N == 0) {` `        ``return` `1;` `    ``}` `    ``long` `long` `int` `sum = 0;` `    ``for` `(``long` `long` `int` `i = 1; i <= N; i++) {` `        ``long` `long` `int` `fact = 1;` `        ``for` `(``long` `long` `int` `j = 1; j <= i; j++) {` `            ``fact *= j;` `            ``fact %= 10; ``// to prevent overflow` `        ``}` `        ``sum += fact;` `        ``sum %= 10; ``// to prevent overflow` `    ``}` `    ``return` `sum;` `}`   `int` `main() {` `    ``long` `long` `int` `N = 1;`   `    ``for` `(N = 0; N <= 10; N++)` `        ``cout << ``"For N = "` `<< N` `             ``<< ``" : "` `<< get_unit_digit(N)` `             ``<< endl;`   `    ``return` `0;` `}`

Output

```For N = 0 : 1
For N = 1 : 1
For N = 2 : 3
For N = 3 : 9
For N = 4 : 3
For N = 5 : 3
For N = 6 : 3
For N = 7 : 3
For N = 8 : 3
For N = 9 : 3
For N = 10 : 3
```

Time Complexity: O(N ^ 2)

Auxiliary Space: O(1)

Efficient Approach: In this approach, only unit’s digit of N is to be calculated in the range [1, 5], because:
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
so on.
As 5!=120, and factorial of number greater than 5 have trailing zeros. So, N>=5 doesn’t contribute in unit place while doing sum.
Therefore:

```if (n < 5)
ans = (1 ! + 2 ! +..+ n !) % 10;
else
ans = (1 ! + 2 ! + 3 ! + 4 !) % 10;

Note : We know (1! + 2! + 3! + 4!) % 10 = 3
So we always return 3 when n is greater
than 4.```

Algorithm :

Step 1: Start
Step 2: Start a static function called get_unit_digit which take integer value as its parameter called N.
Step 3: Now inside the above function set some conditions:
a. If N is 0 or 1, return 1.
b. If N is 2, return 3.
c. If N is 3, return 9.
d. If N is greater than or equal to 4, return 3.
Step 4: End

Below is the implementation of the efficient approach:

## C++

 `// C++ program to find the unit place digit` `// of the first N natural numbers factorials` `#include ` `using` `namespace` `std;`   `// Function to find the unit's place digit` `int` `get_unit_digit(``long` `long` `int` `N)` `{`   `    ``// Let us write for cases when` `    ``// N is smaller than or equal` `    ``// to 4.` `    ``if` `(N == 0 || N == 1)` `       ``return` `1;` `    ``else` `if` `(N == 2)` `       ``return` `3;` `    ``else`  `if` `(N == 3)` `       ``return` `9;`   `    ``// We know following` `    ``// (1! + 2! + 3! + 4!) % 10 = 3` `    ``else` `// (N >= 4) ` `       ``return` `3;` `}`   `// Driver code` `int` `main()` `{` `    ``long` `long` `int` `N = 1;`   `    ``for` `(N = 0; N <= 10; N++)` `        ``cout << ``"For N = "` `<< N` `             ``<< ``" : "` `<< get_unit_digit(N)` `             ``<< endl;`   `    ``return` `0;` `}`

## Java

 `// Java  program to find the unit place digit ` `// of the first N natural numbers factorials`   `import` `java.io.*;`   `class` `GFG {` `    `  `    `  `// Function to find the unit's place digit ` `static` `int` `get_unit_digit(  ``int` `N) ` `{ `   `    ``// Let us write for cases when ` `    ``// N is smaller than or equal ` `    ``// to 4. ` `    ``if` `(N == ``0` `|| N == ``1``) ` `    ``return` `1``; ` `    ``else` `if` `(N == ``2``) ` `    ``return` `3``; ` `    ``else` `if` `(N == ``3``) ` `    ``return` `9``; `   `    ``// We know following ` `    ``// (1! + 2! + 3! + 4!) % 10 = 3 ` `    ``else` `// (N >= 4) ` `    ``return` `3``; ` `} `   `// Driver code ` `    `  `    ``public` `static` `void` `main (String[] args) {` `        `  `      ``int` `N = ``1``; `   `    ``for` `(N = ``0``; N <= ``10``; N++) ` `            ``System.out.println (``"For N = "` `+ N ` `            ``+ ``" : "` `+ get_unit_digit(N)); ` `    ``}` `}` `//This Code is Contributed by ajit`

## Python3

 `# Python3 program to find the unit` `# place digit of the first N natural` `# numbers factorials `   `# Function to find the unit's place digit` `def` `get_unit_digit(N):` `    `  `    ``# Let us write for cases when ` `    ``# N is smaller than or equal ` `    ``# to 4. ` `    ``if` `(N ``=``=` `0` `or` `N ``=``=` `1``):` `        ``return` `1` `    ``elif` `(N ``=``=` `2``):` `        ``return` `3` `    ``elif``(N ``=``=` `3``):` `        ``return` `9` `        `  `    ``# We know following ` `    ``# (1! + 2! + 3! + 4!) % 10 = 3` `    ``else``:` `        ``return` `3`   `# Driver code` `N ``=` `1` `for` `N ``in` `range``(``11``):` `    ``print``(``"For N = "``, N, ``":"``,` `        ``get_unit_digit(N), sep ``=` `' '``)`   `# This code is contributed ` `# by sahilshelangia`

## C#

 `// C# program to find the unit ` `// place digit of the first N ` `// natural numbers factorials` `using` `System;`   `class` `GFG` `{` `    `  `// Function to find the unit's` `// place digit ` `static` `int` `get_unit_digit( ``int` `N) ` `{ `   `    ``// Let us write for cases when ` `    ``// N is smaller than or equal ` `    ``// to 4. ` `    ``if` `(N == 0 || N == 1) ` `    ``return` `1; ` `    ``else` `if` `(N == 2) ` `    ``return` `3; ` `    ``else` `if` `(N == 3) ` `    ``return` `9; `   `    ``// We know following ` `    ``// (1! + 2! + 3! + 4!) % 10 = 3 ` `    ``else` `// (N >= 4) ` `    ``return` `3; ` `} `   `// Driver code ` `static` `public` `void` `Main ()` `{` `    ``int` `N = 1; `   `    ``for` `(N = 0; N <= 10; N++) ` `        ``Console.WriteLine (``"For N = "` `+ N +` `                ``" : "` `+ get_unit_digit(N)); ` `}` `}`   `// This Code is Contributed by akt_mit`

## PHP

 `= 4) ` `        ``return` `3;` `}`   `// Driver code` `\$N` `= 1;`   `for` `(``\$N` `= 0; ``\$N` `<= 10; ``\$N``++)` `    ``echo` `"For N = "` `. ``\$N``.` `         ``" : "` `. get_unit_digit(``\$N``) . ``"\n"``;`   `// This code is contributed ` `// by ChitraNayal` `?>`

## Javascript

 ``

Output

```For N = 0 : 1
For N = 1 : 1
For N = 2 : 3
For N = 3 : 9
For N = 4 : 3
For N = 5 : 3
For N = 6 : 3
For N = 7 : 3
For N = 8 : 3
For N = 9 : 3
For N = 10 : 3
```

Time Complexity: O(1)

Auxiliary Space: O(1)

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