Find the total Number of Digits in (N!)N

Given a number N. The task is to find the total Number of Digits in .

Examples:

```Input: N = 3
Output: 3
If N=3, (3!)3=216,
So the count of digits is 3

Input: N = 4
Output: 6
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

```As we know,
log(a*b) = log(a) + log(b)

Consider,
X = log(N!) = log(1*2*3....... * N)
= log(1)+log(2)+........ +log(N)
```

Now, we know that the floor value of log base 10 increased by 1, of any number, gives the number of digits present in that number. That is, number of digits in a number say N will be floor(log10N) + 1.

Therefore, number of digit in will be:

```floor(log())+1
= floor(N*log10(N!)) + 1
= floor(N*X) + 1.
```

Below is the implementation of the above approach:

C++

 `// C++ program to find the total ` `// Number of Digits in (N!)^N ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the total ` `// Number of Digits in (N!)^N ` `int` `CountDigits(``int` `n) ` `{ ` `    ``if` `(n == 1) ` `        ``return` `1; ` ` `  `    ``double` `sum = 0; ` ` `  `    ``// Finding X ` `    ``for` `(``int` `i = 2; i <= n; ++i) { ` `        ``sum += (``double``)``log``(i) / (``double``)``log``(10); ` `    ``} ` ` `  `    ``// Calculating N*X ` `    ``sum *= (``double``)n; ` ` `  `    ``// Floor(N*X) + 1 ` `    ``return` `ceil``(sum); ``// equivalent to floor(sum) + 1 ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 5; ` ` `  `    ``cout << CountDigits(N); ` ` `  `    ``return` `0; ` `} `

Java

 `// Java program to find the total  ` `// Number of Digits in (N!)^N  ` `import` `java.io.*; ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `class` `GFG ` `{ ` `// Function to find the total  ` `// Number of Digits in (N!)^N  ` `public` `double` `CountDigits(``int` `n)  ` `{  ` `    ``if` `(n == ``1``)  ` `        ``return` `1``;  ` ` `  `    ``double` `sum = ``0``;  ` ` `  `    ``// Finding X  ` `    ``for` `(``int` `i = ``2``; i <= n; ++i) ` `    ``{  ` `        ``sum += ((``double``)Math.log(i) /  ` `                ``(``double``)Math.log(``10``));  ` `    ``}  ` ` `  `    ``// Calculating N*X  ` `    ``sum *= n;  ` ` `  `    ``// Floor(N*X) + 1  ` `    ``// equivalent to floor(sum) + 1  ` `    ``return` `Math.ceil(sum);  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[])  ` `{  ` `    ``GFG g = ``new` `GFG(); ` `    ``int` `N = ``5``;  ` `    ``System.out.println(g.CountDigits(N));  ` `} ` `} ` ` `  `// This code is contributed  ` `// by Akanksha Rai(Abby_akku) `

Python3

 `# Python3 program to find the total  ` `# Number of Digits in (N!)^N ` ` `  `import` `math as ma ` `def` `CountDigits(n): ` ` `  `    ``if``(n``=``=``1``): ` `        ``return` `1` `    ``sum``=``0` ` `  `    ``# Finding X ` `    ``for` `i ``in` `range``(``2``,n``+``1``): ` `        ``sum``+``=``ma.log(i,``10``) ` ` `  `    ``# Calculating N*X  ` `    ``sum``*``=``n ` ` `  `    ``# Floor(N*X)+1 ` `    ``#equivalent to floor(sum) + 1 ` `    ``return` `ma.ceil(``sum``)  ` ` `  `# Driver code ` `if` `__name__``=``=``'__main__'``: ` `    ``N``=``5` `    ``print``(CountDigits(N)) ` ` `  `# This code is contributed by  ` `# Indrajit Sinha. `

C#

 `// C# program to find the total  ` `// Number of Digits in (N!)^N  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `// Function to find the total  ` `// Number of Digits in (N!)^N  ` `public` `double` `CountDigits(``int` `n)  ` `{  ` `    ``if` `(n == 1)  ` `        ``return` `1;  ` ` `  `    ``double` `sum = 0;  ` ` `  `    ``// Finding X  ` `    ``for` `(``int` `i = 2; i <= n; ++i) ` `    ``{  ` `        ``sum += ((``double``)Math.Log(i) /  ` `                ``(``double``)Math.Log(10));  ` `    ``}  ` ` `  `    ``// Calculating N*X  ` `    ``sum *= n;  ` ` `  `    ``// Floor(N*X) + 1  ` `    ``// equivalent to floor(sum) + 1  ` `    ``return` `Math.Ceiling(sum);  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main()  ` `{  ` `    ``GFG g = ``new` `GFG(); ` `    ``int` `N = 5;  ` `    ``Console.WriteLine(g.CountDigits(N));  ` `} ` `} ` ` `  `// This code is contributed  ` `// by SoumikMondal `

PHP

 ` `

Output:

```11
```

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