# Finding number of digits in n’th Fibonacci number

Given a number n, find number of digits in n’th Fibonacci Numbers. First few Fibinacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ….

Examples:

```Input : n = 6
Output : 1
6'th Fibonacci number is 8 and it has
1 digit.

Input : n = 12
Output : 3
12'th Fibonacci number is 144 and it has
3 digits.
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A simple solution is to find n’th Fibonacci Number and then count number of digits in it. This solution may lead to overflow problems for large values of n.

A direct way is to count number of digits in the nth Fibonacci number using below Binet’s Formula.

```fib(n) = (Φn - Ψ-n) / √5
where
Φ = (1 + √5) / 2
Ψ = (1 - √5) / 2

The above formula can be simplified,
fib(n) = round(Φn / ? 5)
Here round function indicates nearest integer.

Count of digits in Fib(n) = Log10Fib(n)
= Log10(Φn / √5)
= n*Log10(Φ) - Log10√5
= n*Log10(Φ) - (Log105)/2
```

As mentioned in this G-Fact, this formula doesn’t seem to work and produce correct Fibonacci numbers due to limitations of floating point arithmetic. However, it looks viable to use this formula to find count of digits in n’th Fibonacci number.

Below is the implementation of above idea :

## C++

 `/* C++ program to find number of digits in nth ` `   ``Fibonacci number */` `#include ` `using` `namespace` `std; ` ` `  `// This function returns the number of digits ` `// in nth Fibonacci number after ceiling it ` `// Formula used (n * log(phi) - (log 5) / 2) ` `long` `long` `numberOfDigits(``long` `long` `n) ` `{ ` `    ``if` `(n == 1) ` `        ``return` `1; ` ` `  `    ``// using phi = 1.6180339887498948 ` `    ``long` `double` `d = (n * ``log10``(1.6180339887498948)) - ` `                    ``((``log10``(5)) / 2); ` ` `  `    ``return` `ceil``(d); ` `} ` ` `  `// Driver program to test the above function ` `int` `main() ` `{ ` `    ``long` `long` `i; ` `    ``for` `(i = 1; i <= 10; i++) ` `    ``cout << ``"Number of Digits in F("` `         ``<< i <<``") - "` `         ``<< numberOfDigits(i) << ``"n"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program  to find number of digits in nth ` `// Fibonacci number ` ` `  `class` `GFG  ` `{ ` `    ``// This function returns the number of digits ` `    ``// in nth Fibonacci number after ceiling it ` `    ``// Formula used (n * log(phi) - (log 5) / 2) ` `    ``static` `double` `numberOfDigits(``double` `n) ` `    ``{ ` `        ``if` `(n == ``1``) ` `            ``return` `1``; ` `     `  `        ``// using phi = 1.6180339887498948 ` `        ``double` `d = (n * Math.log10(``1.6180339887498948``)) - ` `                   ``((Math.log10(``5``)) / ``2``); ` `     `  `        ``return` `Math.ceil(d); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``double` `i; ` `        ``for` `(i = ``1``; i <= ``10``; i++) ` `        ``System.out.println(``"Number of Digits in F("``+i+``") - "`  `                           ``+numberOfDigits(i)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python program to find  ` `# number of digits in nth  ` `# Fibonacci number ` `import` `math ` ` `  `# storing value of  ` `# golden ratio aka phi ` `phi ``=` `(``1` `+` `5``*``*``.``5``) ``/` `2` ` `  `# function to find number  ` `# of digits in F(n) This  ` `# function returns the number  ` `# of digitsin nth Fibonacci  ` `# number after ceiling it ` `# Formula used (n * log(phi) -  ` `# (log 5) / 2) ` `def` `numberOfDig (n) : ` `    ``if` `n ``=``=` `1` `: ` `        ``return` `1` `    ``return` `math.ceil((n ``*` `math.log10(phi) ``-`  `                      ``.``5` `*` `math.log10(``5``))) ` ` `  `/``/` `Driver Code ` `for` `i ``in` `range``(``1``, ``11``) : ` `    ``print``(``"Number of Digits in F("` `+`  `                   ``str``(i) ``+` `") - "` `+`  `                ``str``(numberOfDig(i))) ` ` `  `# This code is contributed by SujanDutta `

## C#

 `// C# program to find number of  ` `// digits in nth Fibonacci number ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// This function returns the number of digits ` `    ``// in nth Fibonacci number after ceiling it ` `    ``// Formula used (n * log(phi) - (log 5) / 2) ` `    ``static` `double` `numberOfDigits(``double` `n) ` `    ``{ ` `        ``if` `(n == 1) ` `            ``return` `1; ` `     `  `        ``// using phi = 1.6180339887498948 ` `        ``double` `d = (n * Math.Log10(1.6180339887498948)) - ` `                   ``((Math.Log10(5)) / 2); ` `     `  `        ``return` `Math.Ceiling(d); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``double` `i; ` `        ``for` `(i = 1; i <= 10; i++) ` `        ``Console.WriteLine(``"Number of Digits in F("``+ i +``") - "` `                           ``+ numberOfDigits(i)); ` `    ``} ` `} ` ` `  `// This code is contributed by Nitin Mittal. `

## PHP

 ` `

Output:

```Number of Digits in F(1) - 1
Number of Digits in F(2) - 1
Number of Digits in F(3) - 1
Number of Digits in F(4) - 1
Number of Digits in F(5) - 1
Number of Digits in F(6) - 1
Number of Digits in F(7) - 2
Number of Digits in F(8) - 2
Number of Digits in F(9) - 2
Number of Digits in F(10) - 2
```

This article is contributed by Ayush Khanduri. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.