# Numbers in a Range with given Digital Root

Given an integer K and a range of consecutive numbers [L, R]. The task is to count the numbers from the given range which have digital root as K (1 ≤ K ≤ 9). Digital root is sum of digits of a number until it becomes a single digit number. For example, 256 -> 2 + 5 + 6 = 13 -> 1 + 3 = 4.

Examples:

Input: L = 10, R = 22, K = 3
Output: 2
12 and 21 are the only numbers from the range whose digit sum is 3.

Input: L = 100, R = 200, K = 5
Output: 11

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

Approach:

• First thing is to note that for any number Sum of Digits is equal to Number % 9. If remainder is 0, then sum of digits is 9.
• So if K = 9, then replace K with 0.
• Task, now is to find count of numbers in range L to R with modulo 9 equal to K.
• Divide the entire range into the maximum possible groups of 9 starting with L (TotalRange / 9), since in each range there will be exactly one number with modulo 9 equal to K.
• Loop over rest number of elements from R to R – count of rest elements, and check if any number satisfies the condition.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `// Function to return the count ` `// of required numbers ` `int` `countNumbers(``int` `L, ``int` `R, ``int` `K) ` `{ ` `    ``if` `(K == 9) ` `        ``K = 0; ` ` `  `    ``// Count of numbers present ` `    ``// in given range ` `    ``int` `totalnumbers = R - L + 1; ` ` `  `    ``// Number of groups of 9 elements ` `    ``// starting from L ` `    ``int` `factor9 = totalnumbers / 9; ` ` `  `    ``// Left over elements not covered ` `    ``// in factor 9 ` `    ``int` `rem = totalnumbers % 9; ` ` `  `    ``// One Number in each group of 9 ` `    ``int` `ans = factor9; ` ` `  `    ``// To check if any number in rem ` `    ``// satisfy the property ` `    ``for` `(``int` `i = R; i > R - rem; i--) { ` `        ``int` `rem1 = i % 9; ` `        ``if` `(rem1 == K) ` `            ``ans++; ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `L = 10; ` `    ``int` `R = 22; ` `    ``int` `K = 3; ` `    ``cout << countNumbers(L, R, K); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` ` `  `class` `GFG { ` ` `  `// Function to return the count ` `// of required numbers ` `    ``static` `int` `countNumbers(``int` `L, ``int` `R, ``int` `K) { ` `        ``if` `(K == ``9``) { ` `            ``K = ``0``; ` `        ``} ` ` `  `        ``// Count of numbers present ` `        ``// in given range ` `        ``int` `totalnumbers = R - L + ``1``; ` ` `  `        ``// Number of groups of 9 elements ` `        ``// starting from L ` `        ``int` `factor9 = totalnumbers / ``9``; ` ` `  `        ``// Left over elements not covered ` `        ``// in factor 9 ` `        ``int` `rem = totalnumbers % ``9``; ` ` `  `        ``// One Number in each group of 9 ` `        ``int` `ans = factor9; ` ` `  `        ``// To check if any number in rem ` `        ``// satisfy the property ` `        ``for` `(``int` `i = R; i > R - rem; i--) { ` `            ``int` `rem1 = i % ``9``; ` `            ``if` `(rem1 == K) { ` `                ``ans++; ` `            ``} ` `        ``} ` ` `  `        ``return` `ans; ` `    ``} ` ` `  `// Driver code ` `    ``public` `static` `void` `main(String[] args) { ` `        ``int` `L = ``10``; ` `        ``int` `R = ``22``; ` `        ``int` `K = ``3``; ` `        ``System.out.println(countNumbers(L, R, K)); ` `    ``} ` `} ` `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count ` `# of required numbers ` `def` `countNumbers(L, R, K): ` ` `  `    ``if` `(K ``=``=` `9``): ` `        ``K ``=` `0` ` `  `    ``# Count of numbers present ` `    ``# in given range ` `    ``totalnumbers ``=` `R ``-` `L ``+` `1` ` `  `    ``# Number of groups of 9 elements ` `    ``# starting from L ` `    ``factor9 ``=` `totalnumbers ``/``/` `9` ` `  `    ``# Left over elements not covered ` `    ``# in factor 9 ` `    ``rem ``=` `totalnumbers ``%` `9` ` `  `    ``# One Number in each group of 9 ` `    ``ans ``=` `factor9 ` ` `  `    ``# To check if any number in rem ` `    ``# satisfy the property ` `    ``for` `i ``in` `range``(R, R ``-` `rem, ``-``1``): ` `        ``rem1 ``=` `i ``%` `9` `        ``if` `(rem1 ``=``=` `K): ` `            ``ans ``+``=` `1` `     `  `    ``return` `ans ` ` `  `# Driver code ` `L ``=` `10` `R ``=` `22` `K ``=` `3` `print``(countNumbers(L, R, K)) ` ` `  `# This code is contributed  ` `# by mohit kumar `

## C#

 `// C# implementation of the approach ` `using` `System ; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to return the count ` `    ``// of required numbers ` `    ``static` `int` `countNumbers(``int` `L, ``int` `R, ``int` `K) ` `    ``{ ` `        ``if` `(K == 9)  ` `        ``{ ` `            ``K = 0; ` `        ``} ` ` `  `        ``// Count of numbers present ` `        ``// in given range ` `        ``int` `totalnumbers = R - L + 1; ` ` `  `        ``// Number of groups of 9 elements ` `        ``// starting from L ` `        ``int` `factor9 = totalnumbers / 9; ` ` `  `        ``// Left over elements not covered ` `        ``// in factor 9 ` `        ``int` `rem = totalnumbers % 9; ` ` `  `        ``// One Number in each group of 9 ` `        ``int` `ans = factor9; ` ` `  `        ``// To check if any number in rem ` `        ``// satisfy the property ` `        ``for` `(``int` `i = R; i > R - rem; i--) ` `        ``{ ` `            ``int` `rem1 = i % 9; ` `            ``if` `(rem1 == K) ` `            ``{ ` `                ``ans++; ` `            ``} ` `        ``} ` ` `  `        ``return` `ans; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `L = 10; ` `        ``int` `R = 22; ` `        ``int` `K = 3; ` `         `  `        ``Console.WriteLine(countNumbers(L, R, K)); ` `    ``} ` `} ` ` `  `/* This code is contributed by Ryuga */`

## PHP

 ` ``\$R` `- ``\$rem``; ``\$i``--)  ` `    ``{ ` `        ``\$rem1` `= ``\$i` `% 9; ` `        ``if` `(``\$rem1` `== ``\$K``) ` `            ``\$ans``++; ` `    ``} ` ` `  `    ``return` `\$ans``; ` `} ` ` `  `// Driver code ` `\$L` `= 10; ` `\$R` `= 22; ` `\$K` `= 3; ` `echo` `countNumbers(``\$L``, ``\$R``, ``\$K``); ` ` `  `// This code is contributed by Ita_c ` `?> `

Output:

```2
```

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