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# Finding number of digits in n’th Fibonacci number

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

`Input : n = 6Output : 16'th Fibonacci number is 8 and it has 1 digit.Input : n = 12Output : 312'th Fibonacci number is 144 and it has 3 digits.`

Recommended Practice

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) / √5whereΦ = (1 + √5) / 2Ψ = (1 - √5) / 2The 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.`

## Javascript

 ``

## 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

```

Time Complexity: O(1)
Auxiliary Space: O(1)

### Another Approach(Using fact that Fibonacci numbers are periodic):

Fibonacci sequence is periodic modulo any integer, with period equal to 60 (known as Pisano’s period). This means that we can calculate the nth Fibonacci number modulo 10^k for some large k, and then use the periodicity to compute the number of digits. For example, we can compute F_n modulo 10^10 and count the number of digits:

F_n_mod = F_n % 10**10
digits = floor(log10(F_n_mod)) + 1

Below is the implementation of above approach:

## C++

 `#include``using` `namespace` `std;` `long` `long` `numberOfDigits(``long` `long` `n){``    ``int` `k = 10; ``// module 10^k``    ``int` `phi = (1 + ``sqrt``(5)) / 2; ``//golden ratio``    ` `    ``// compute the n-th Fibonacci number modulo 10^k``    ``int` `a = 0, b = 1;``    ``for` `(``int` `i = 2; i <= n; i++) {``        ``int` `c = (a + b) % ``int``(``pow``(10, k));``        ``a = b;``        ``b = c;``    ``}``    ``int` `F_n_mod = b;` `    ``// compute the number of digits in F_n_mod``    ``int` `digits = 1;``    ``while` `(F_n_mod >= 10) {``        ``F_n_mod /= 10;``        ``digits++;``    ``}``    ``return` `digits;``}` `int` `main(){``    ``long` `long` `i;``    ``for` `(i = 1; i <= 10; i++)``    ``cout << ``"Number of Digits in F("``         ``<< i <<``") - "``         ``<< numberOfDigits(i) << ``"\n"``;``    ``return` `0;``}` `// This code is contributed by Yash Agarwal(yashagarwal2852002)`

## Python3

 `import` `math`  `def` `numberOfDigits(n):``    ``k ``=` `10``    ``# Golden ratio (approximately 1.618033988749895)``    ``phi ``=` `(``1` `+` `math.sqrt(``5``)) ``/` `2` `    ``# Compute the n-th Fibonacci number modulo 10^k``    ``a, b ``=` `0``, ``1``    ``# Start the loop from 2, as we already have F(0) and F(1)``    ``for` `i ``in` `range``(``2``, n ``+` `1``):``        ``c ``=` `(a ``+` `b) ``%` `pow``(``10``, k)``        ``# Update the previous Fibonacci numbers for the next iteration``        ``a ``=` `b``        ``b ``=` `c``    ``F_n_mod ``=` `b` `    ``# Compute the number of digits in F_n_mod``    ``# Initialize the digit counter to 1 (as any number has at least one digit)``    ``digits ``=` `1``    ``# Keep dividing F_n_mod by 10 until it becomes less than 10``    ``while` `F_n_mod >``=` `10``:``        ``F_n_mod ``=` `F_n_mod ``/``/` `10``        ``# Increment the digit counter``        ``digits ``+``=` `1``# Return the number of digits in the n-th Fibonacci number modulo 10^k``    ``return` `digits`  `# Driver code``for` `i ``in` `range``(``1``, ``11``):``    ``# Calculate and print the number of digits in F(i) modulo 10^10``    ``print``(``"Number of Digits in F("` `+` `str``(i) ``+` `") - "` `+` `str``(numberOfDigits(i)))` `# THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)`

## Javascript

 `function` `numberOfDigits(n) {``  ``let k = 10; ``// module 10^k``  ``let phi = (1 + Math.sqrt(5)) / 2; ``// golden ratio` `  ``// compute the n-th Fibonacci number modulo 10^k``  ``let a = 0,``    ``b = 1;``  ``for` `(let i = 2; i <= n; i++) {``    ``let c = (a + b) % Math.pow(10, k);``    ``a = b;``    ``b = c;``  ``}``  ``let F_n_mod = b;` `  ``// compute the number of digits in F_n_mod``  ``let digits = 1;``  ``while` `(F_n_mod >= 10) {``    ``F_n_mod = Math.floor(F_n_mod / 10);``    ``digits++;``  ``}``  ``return` `digits;``}` `// main function``let i;``for` `(i = 1; i <= 10; i++)``  ``console.log(``"Number of Digits in F("` `+ i + ``") - "` `+ numberOfDigits(i));` `// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)`

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

```

Time Complexity: O(nk)

Auxiliary Space: O(1)

References:
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html#section2
https://en.wikipedia.org/wiki/Fibonacci_number
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