# Find Nth positive number whose digital root is X

• Difficulty Level : Medium
• Last Updated : 22 Jun, 2022

Given a number X ( 1<= X <= 9) and a positive number N, find the Nth positive number whose digital root is X.
Digital Root: The digital root of a positive number is obtained by iteratively summing up the digits of a number. On each iteration, the number is replaced by the sum of its digit and the iteration stops when the number is reduced to a single digit number. This single digit number is known as the digital root. For example, the digital root of 65 is 2, because 6 + 5 = 11 and 1 + 1 = 2.

Examples:

Input : X = 3, N = 100
Output : 894
The N-th Number whose digit root is X is 894

Input : X = 7, N = 43
Output : 385

Simple Method: The simple method is to start from 1 and calculate the digital root of every number, whenever we come across a number whose digital root is equal to X, we increase our counter by 1. We will stop our search when our counter is equal to N.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the N-th number whose``// digital root is X` `#include ``using` `namespace` `std;` `// Function to find the digital root of``// a number``int` `findDigitalRoot(``int` `num)``{``    ``int` `sum = INT_MAX, tempNum = num;` `    ``while` `(sum >= 10) {``        ``sum = 0;` `        ``while` `(tempNum > 0) {``            ``sum += tempNum % 10;``            ``tempNum /= 10;``        ``}` `        ``tempNum = sum;``    ``}` `    ``return` `sum;``}` `// Function to find the Nth number with``// digital root as X``void` `findAnswer(``int` `X, ``int` `N)``{``    ``// Counter variable to keep the``    ``// count of valid numbers``    ``int` `counter = 0;` `    ``for` `(``int` `i = 1; counter < N; ++i) {` `        ``// Find digital root``        ``int` `digitalRoot = findDigitalRoot(i);` `        ``// Check if is required answer or not``        ``if` `(digitalRoot == X) {``            ``++counter;``        ``}` `        ``// Print the answer if you have found it``        ``// and breakout of the loop``        ``if` `(counter == N) {``            ``cout << i;` `            ``break``;``        ``}``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `X = 1, N = 3;` `    ``findAnswer(X, N);` `    ``return` `0;``}`

## Java

 `// Java program to find the N-th number whose``// digital root is X``class` `GFG``{` `// Function to find the digital root of``// a number``static` `int` `findDigitalRoot(``int` `num)``{``    ``int` `sum = Integer.MAX_VALUE, tempNum = num;` `    ``while` `(sum >= ``10``)``    ``{``        ``sum = ``0``;` `        ``while` `(tempNum > ``0``)``        ``{``            ``sum += tempNum % ``10``;``            ``tempNum /= ``10``;``        ``}` `        ``tempNum = sum;``    ``}` `    ``return` `sum;``}` `// Function to find the Nth number with``// digital root as X``static` `void` `findAnswer(``int` `X, ``int` `N)``{``    ``// Counter variable to keep the``    ``// count of valid numbers``    ``int` `counter = ``0``;` `    ``for` `(``int` `i = ``1``; counter < N; ++i)``    ``{` `        ``// Find digital root``        ``int` `digitalRoot = findDigitalRoot(i);` `        ``// Check if is required answer or not``        ``if` `(digitalRoot == X)``        ``{``            ``++counter;``        ``}` `        ``// Print the answer if you have found it``        ``// and breakout of the loop``        ``if` `(counter == N)``        ``{``            ``System.out.print( i);` `            ``break``;``        ``}``    ``}``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``int` `X = ``1``, N = ``3``;` `    ``findAnswer(X, N);``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 program to find the N-th number whose``# digital root is X``import` `sys` `# Function to find the digital root of``# a number``def` `findDigitalRoot(num):``    ``sum` `=` `sys.maxsize;``    ``tempNum ``=` `num;` `    ``while` `(``sum` `>``=` `10``):``        ``sum` `=` `0``;` `        ``while` `(tempNum > ``0``):``            ``sum` `+``=` `tempNum ``%` `10``;``            ``tempNum ``/``/``=` `10``;` `        ``tempNum ``=` `sum``;` `    ``return` `sum``;` `# Function to find the Nth number with``# digital root as X``def` `findAnswer(X, N):``    ` `    ``# Counter variable to keep the``    ``# count of valid numbers``    ``counter ``=` `0``;``    ``i ``=` `0``;``    ``while` `(counter < N):``        ``i ``+``=` `1``;``        ` `        ``# Find digital root``        ``digitalRoot ``=` `findDigitalRoot(i);``        ` `        ``# Check if is required answer or not``        ``if` `(digitalRoot ``=``=` `X):``            ``counter ``+``=` `1``;``        ` `        ``# Print the answer if you have found it``        ``# and breakout of the loop``        ``if` `(counter ``=``=` `N):``            ``print``(i);``            ``break``;` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``X ``=` `1``;``    ``N ``=` `3``;` `    ``findAnswer(X, N);` `# This code is contributed by 29AjayKumar`

## C#

 `// C# program to find the N-th number whose``// digital root is X``using` `System;` `class` `GFG``{` `// Function to find the digital root of``// a number``static` `int` `findDigitalRoot(``int` `num)``{``    ``int` `sum = ``int``.MaxValue, tempNum = num;` `    ``while` `(sum >= 10)``    ``{``        ``sum = 0;` `        ``while` `(tempNum > 0)``        ``{``            ``sum += tempNum % 10;``            ``tempNum /= 10;``        ``}` `        ``tempNum = sum;``    ``}` `    ``return` `sum;``}` `// Function to find the Nth number with``// digital root as X``static` `void` `findAnswer(``int` `X, ``int` `N)``{``    ``// Counter variable to keep the``    ``// count of valid numbers``    ``int` `counter = 0;` `    ``for` `(``int` `i = 1; counter < N; ++i)``    ``{` `        ``// Find digital root``        ``int` `digitalRoot = findDigitalRoot(i);` `        ``// Check if is required answer or not``        ``if` `(digitalRoot == X)``        ``{``            ``++counter;``        ``}` `        ``// Print the answer if you have found it``        ``// and breakout of the loop``        ``if` `(counter == N)``        ``{``            ``Console.Write( i);` `            ``break``;``        ``}``    ``}``}` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ``int` `X = 1, N = 3;` `    ``findAnswer(X, N);``}``}` `// This code has been contributed by 29AjayKumar`

## PHP

 `= 10)``    ``{``        ``\$sum` `= 0;` `        ``while` `(``\$tempNum` `> 0)``        ``{``            ``\$sum` `+= ``\$tempNum` `% 10;``            ``\$tempNum` `/= 10;``        ``}` `        ``\$tempNum` `= ``\$sum``;``    ``}` `    ``return` `\$sum``;``}` `// Function to find the Nth number ``// with digital root as X``function` `findAnswer(``\$X``, ``\$N``)``{``    ` `    ``// Counter variable to keep the``    ``// count of valid numbers``    ``\$counter` `= 0;` `    ``for` `(``\$i` `= 1; ``\$counter` `< ``\$N``; ++``\$i``)``    ``{` `        ``// Find digital root``        ``\$digitalRoot` `= findDigitalRoot(``\$i``);` `        ``// Check if is required answer or not``        ``if` `(``\$digitalRoot` `== ``\$X``)``        ``{``            ``++``\$counter``;``        ``}` `        ``// Print the answer if you have found``        ``// it and breakout of the loop``        ``if` `(``\$counter` `== ``\$N``)``        ``{``            ``echo``( ``\$i``);` `            ``break``;``        ``}``    ``}``}` `// Driver Code``\$X` `= 1; ``\$N` `= 3;` `findAnswer(``\$X``, ``\$N``);` `// This code is contributed by Code_Mech.`

## Javascript

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Output:

`19`

Efficient Approach: We can find digital root of a number K directly using the formula:

`digitalRoot(k) = (k - 1)mod 9 +1`

From this we can find the N-th number whose digital root is K as,

`Nth number = (N - 1)*9 + K`

Below is the implementation of the above approach:

## C++

 `// C++ program to find the N-th number with``// digital root as X` `#include ``using` `namespace` `std;` `// Function to find the N-th number with``// digital root as X``int` `findAnswer(``int` `X, ``int` `N)``{``    ``return` `(N - 1) * 9 + X;``}` `// Driver Code``int` `main()``{``    ``int` `X = 7, N = 43;` `    ``cout << findAnswer(X, N);` `    ``return` `0;``}`

## Java

 `// Java program to find the N-th number with``// digital root as X` `class` `GfG``{` `    ``// Function to find the N-th number with``    ``// digital root as X``    ``static` `int` `findAnswer(``int` `X, ``int` `N)``    ``{``        ``return` `(N - ``1``) * ``9` `+ X;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `X = ``7``, N = ``43``;``        ``System.out.println(findAnswer(X, N));``    ``}``}` `// This code contributed by Rajput-Ji`

## Python3

 `# Python3 program to find the N-th ``# number with digital root as X` `# Function to find the N-th number``# with digital root as X``def` `findAnswer(X, N):` `    ``return` `(N ``-` `1``) ``*` `9` `+` `X;` `# Driver Code``X ``=` `7``;``N ``=` `43``;``print``(findAnswer(X, N));` `# This code is contributed by mits`

## C#

 `// C# program to find the N-th number``// with digital root as X``using` `System;` `class` `GFG``{` `// Function to find the N-th``// number with digital root as X``static` `int` `findAnswer(``int` `X, ``int` `N)``{``    ``return` `(N - 1) * 9 + X;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `X = 7, N = 43;``    ``Console.WriteLine(findAnswer(X, N));``}``}` `// This code contributed by Ryuga`

## PHP

 ``

## Javascript

 ``

Output:

`385`

Time Complexity: O(1)

Auxiliary Space: O(1)

My Personal Notes arrow_drop_up