# Program to find N-th term of series 3, 5, 33, 35, 53….

• Last Updated : 26 May, 2022

Given a series of numbers composed only of digits 3 and 5. The first few numbers in the series are:

3, 5, 33, 35, 53, 55, …..

Given a number N. The task is to find the n-th number in the given series.
Examples

```Input : N = 2
Output : 5

Input : N = 5
Output : 53```

The idea is based on the fact that the value of the last digit alternates in the series. For example, if the last digit of ith number is 3, then the last digit of (i-1)th and (i+1)th numbers must be 5.
Create an array of size (n+1) and push 3 and 5(These two are always first two elements of series) to it. For more elements check,

```1) If i is odd,
arr[i] = arr[i/2]*10 + 3;
2) If it is even,
arr[i] = arr[(i/2)-1]*10 + 5;
At last return arr[n].```

Below is the implementation of the above idea:

## C++

 `// C++ program to find n-th number in a series``// made of digits 3 and 5` `#include ``using` `namespace` `std;` `// Function to find n-th number in series``// made of 3 and 5``int` `printNthElement(``int` `n)``{``    ``// create an array of size (n+1)``    ``int` `arr[n + 1];``    ``arr = 3;``    ``arr = 5;` `    ``for` `(``int` `i = 3; i <= n; i++) {``        ``// If i is odd``        ``if` `(i % 2 != 0)``            ``arr[i] = arr[i / 2] * 10 + 3;``        ``else``            ``arr[i] = arr[(i / 2) - 1] * 10 + 5;``    ``}``    ``return` `arr[n];``}` `// Driver code``int` `main()``{``    ``int` `n = 6;` `    ``cout << printNthElement(n);` `    ``return` `0;``}`

## C

 `// C program to find n-th number in a series``// made of digits 3 and 5``#include ` `// Function to find n-th number in series``// made of 3 and 5``int` `printNthElement(``int` `n)``{``    ``// create an array of size (n+1)``    ``int` `arr[n + 1];``    ``arr = 3;``    ``arr = 5;` `    ``for` `(``int` `i = 3; i <= n; i++) {``        ``// If i is odd``        ``if` `(i % 2 != 0)``            ``arr[i] = arr[i / 2] * 10 + 3;``        ``else``            ``arr[i] = arr[(i / 2) - 1] * 10 + 5;``    ``}``    ``return` `arr[n];``}` `// Driver code``int` `main()``{``    ``int` `n = 6;` `    ``printf``(``"%d"``,printNthElement(n));` `    ``return` `0;``}` `// This code is contributed by kothavvsaakash.`

## Java

 `// Java program to find n-th number in a series``// made of digits 3 and 5` `class` `FindNth {``    ``// Function to find n-th number in series``    ``// made of 3 and 5``    ``static` `int` `printNthElement(``int` `n)``    ``{``        ``// create an array of size (n+1)``        ``int` `arr[] = ``new` `int``[n + ``1``];``        ``arr[``1``] = ``3``;``        ``arr[``2``] = ``5``;` `        ``for` `(``int` `i = ``3``; i <= n; i++) {``            ``// If i is odd``            ``if` `(i % ``2` `!= ``0``)``                ``arr[i] = arr[i / ``2``] * ``10` `+ ``3``;``            ``else``                ``arr[i] = arr[(i / ``2``) - ``1``] * ``10` `+ ``5``;``        ``}``        ``return` `arr[n];``    ``}` `    ``// main function``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``6``;` `        ``System.out.println(printNthElement(n));``    ``}``}`

## Python3

 `# Python3 program to find n-th number ``# in a series made of digits 3 and 5``  ` `# Return n-th number in series made ``# of 3 and 5``def` `printNthElement(n) :``      ` `    ``# create an array of size (n + 1)``    ``arr ``=``[``0``] ``*` `(n ``+` `1``);``    ``arr[``1``] ``=` `3``    ``arr[``2``] ``=` `5``  ` `    ``for` `i ``in` `range``(``3``, n ``+` `1``) :``        ``# If i is odd``        ``if` `(i ``%` `2` `!``=` `0``) :``            ``arr[i] ``=` `arr[i ``/``/` `2``] ``*` `10` `+` `3``        ``else` `:``            ``arr[i] ``=` `arr[(i ``/``/` `2``) ``-` `1``] ``*` `10` `+` `5``      ` `    ``return` `arr[n]``      ` `# Driver code``n ``=` `6``print``(printNthElement(n))`

## C#

 `// C# program to find n-th number``// in a series made of digits 3 and 5``using` `System;` `class` `GFG``{``// Function to find n-th number``// in series made of 3 and 5``static` `int` `printNthElement(``int` `n)``{``    ``// create an array of size (n+1)``    ``int` `[] arr = ``new` `int``[n + 1];``    ``arr = 3;``    ``arr = 5;` `    ``for` `(``int` `i = 3; i <= n; i++)``    ``{``        ``// If i is odd``        ``if` `(i % 2 != 0)``            ``arr[i] = arr[i / 2] * 10 + 3;``        ``else``            ``arr[i] = arr[(i / 2) - 1] * 10 + 5;``    ``}``    ``return` `arr[n];``}` `// Driver Code``static` `void` `Main()``{``    ``int` `n = 6;` `    ``Console.WriteLine(printNthElement(n));``}``}` `// This code is contributed by ANKITRAI1`

## PHP

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## Javascript

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Output:

`55`

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