n-th number whose sum of digits is ten

Given an integer value n, find out the n-th positive integer whose sum is 10.

Examples:

Input : n = 2
Output : 28
The first number with sum of digits as
10 is 19. Second number is 28.

Input : 15
Output : 154

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

Method 1 (Simple):
We traverse through all numbers. For every number, we find the sum of digits. We stop when we find the n-th number with sum of digits as 10.

C++

 // Simple CPP program to find n-th number // with sum of digits as 10. #include using namespace std;    int findNth(int n) {     int count = 0;        for (int curr = 1;; curr++) {            // Find sum of digits in current no.         int sum = 0;         for (int x = curr; x > 0; x = x / 10)             sum = sum + x % 10;            // If sum is 10, we increment count         if (sum == 10)             count++;            // If count becomes n, we return current         // number.         if (count == n)             return curr;     }     return -1; }    int main() {     printf("%d\n", findNth(5));     return 0; }

Java

 // Java program to find n-th number // with sum of digits as 10.    import java.util.*; import java.lang.*;    public class GFG {     public static int findNth(int n)     {         int count = 0;         for (int curr = 1;; curr++) {                // Find sum of digits in current no.             int sum = 0;             for (int x = curr; x > 0; x = x / 10)                 sum = sum + x % 10;                // If sum is 10, we increment count             if (sum == 10)                 count++;                // If count becomes n, we return current             // number.             if (count == n)                 return curr;         }     }        public static void main(String[] args)     {         System.out.print(findNth(5));     } }    // Contributed by _omg

Python3

 # Python3 program to find n-th number # with sum of digits as 10. import itertools    # function to find required number def findNth(n):        count = 0        for curr in itertools.count():         # Find sum of digits in current no.         sum = 0         x = curr         while(x):             sum = sum + x % 10             x = x // 10            # If sum is 10, we increment count         if (sum == 10):             count = count + 1            # If count becomes n, we return current         # number.         if (count == n):             return curr        return -1    # Driver program if __name__=='__main__':     print(findNth(5))    # This code is contributed by # Sanjit_Prasad

C#

 // C# program to find n-th number // with sum of digits as 10. using System;    class GFG {     public static int findNth(int n)     {         int count = 0;         for (int curr = 1;; curr++) {                // Find sum of digits in current no.             int sum = 0;             for (int x = curr; x > 0; x = x / 10)                 sum = sum + x % 10;                // If sum is 10, we increment count             if (sum == 10)                 count++;                // If count becomes n, we             // return current number.             if (count == n)                 return curr;         }     }        // Driver Code     static public void Main()     {         Console.WriteLine(findNth(5));     } }    // This code is contributed // by Sach_Code

PHP

 0; \$x = \$x / 10)              \$sum = \$sum + \$x % 10;             // If sum is 10, we increment          // count          if (\$sum == 10)              \$count++;             // If count becomes n, we return          // current number.          if (\$count == \$n)              return \$curr;      }      return -1;  }     // Driver Code echo findNth(5);     // This code is contributed by Sach . ?>

Output:

55

Method 2 (Efficient):
If we take a closer look, we can notice that all multiples of 9 are present in arithmetic progression 19, 28, 37, 46, 55, 64, 73, 82, 91, 100, 109, ….
However, there are numbers in above series whose sum of digits is not 10, for example, 100. So instead of checking one by one, we start with 19 and increment by 9.

C++

 // Simple CPP program to find n-th number // with sum of digits as 10. #include using namespace std;    int findNth(int n) {     int count = 0;        for (int curr = 19;; curr += 9) {            // Find sum of digits in current no.         int sum = 0;         for (int x = curr; x > 0; x = x / 10)             sum = sum + x % 10;            // If sum is 10, we increment count         if (sum == 10)             count++;            // If count becomes n, we return current         // number.         if (count == n)             return curr;     }     return -1; }    int main() {     printf("%d\n", findNth(5));     return 0; }

Java

 // Java program to find n-th number // with sum of digits as 10.    import java.util.*; import java.lang.*;    public class GFG {     public static int findNth(int n)     {         int count = 0;            for (int curr = 19;; curr += 9) {                // Find sum of digits in current no.             int sum = 0;             for (int x = curr; x > 0; x = x / 10)                 sum = sum + x % 10;                // If sum is 10, we increment count             if (sum == 10)                 count++;                // If count becomes n, we return current             // number.             if (count == n)                 return curr;         }     }        public static void main(String[] args)     {         System.out.print(findNth(5));     } }    // Contributed by _omg

Python3

 # Python3 program to find n-th  # number with sum of digits as 10.  def findNth(n):      count = 0;            curr = 19;        while (True):             # Find sum of digits in         # current no.          sum = 0;         x = curr;         while (x > 0):             sum = sum + x % 10;             x = int(x / 10);            # If sum is 10, we increment         # count          if (sum == 10):              count+= 1;             # If count becomes n, we return          # current number.          if (count == n):              return curr;                    curr += 9;        return -1;     # Driver Code print(findNth(5));     # This code is contributed  # by mits

C#

 // C# program to find n-th number // with sum of digits as 10. using System;    class GFG {     public static int findNth(int n)     {         int count = 0;            for (int curr = 19;; curr += 9) {                // Find sum of digits in             // current no.             int sum = 0;             for (int x = curr;                  x > 0; x = x / 10)                 sum = sum + x % 10;                // If sum is 10, we increment             // count             if (sum == 10)                 count++;                // If count becomes n, we return             // current number.             if (count == n)                 return curr;         }     }        // Driver Code     static public void Main()     {         Console.WriteLine(findNth(5));     } }    // This code is contributed // by Sach_Code

PHP

 0;              \$x = (int)\$x / 10)              \$sum = \$sum + \$x % 10;             // If sum is 10, we increment         // count          if (\$sum == 10)              \$count++;             // If count becomes n, we return          // current number.          if (\$count == \$n)              return \$curr;      }      return -1;  }     // Driver Code echo findNth(5);     // This code is contributed  // by Sach_Code ?>

Output:

55

Method 3 (Constant Space and Constant Time):

If we see the List closely, we will find out that it is an AP where a = 19 and d = 9. But it has got some outliers too which are in multiples of 10 and starts from 100. So the sequence of outliers is something like – {100, 1000, 10000, …}

Now if we want to find the nth perfect number, it should be (nth number from AP described above + 9 * number of outliers). Now let’s have a look at how to find the number of outliers. Let’s have a close look at the values returned by the Log10 function for a value n and number of outliers till n –

n Log10 Number of Outliers
10 1 0
100 2 1
1000 3 2

So at any point, if we subtract 1 from the log10 of the nth element and cast the result into an integer, we will get the number of outliers.

Following code implements the above approach:

C++

 // CPP program to find n-th number // with sum of digits as 10. #include    using namespace std;    int findNth(int n) {     int nthElement = 19 + (n - 1) * 9;     int outliersCount = (int)log10(nthElement) - 1;        // find the nth perfect number     nthElement += 9 * outliersCount;     return nthElement; }    int main() {     cout << findNth(5) << endl;     return 0; }    // This code is contributed by Rituraj Jain

Java

 // Java program to find n-th number // with sum of digits as 10. import java.lang.Math; class GFG {        public static int findNth(int n)     {         int nthElement = 19 + (n - 1) * 9;         int outliersCount = (int)Math.log10(nthElement) - 1;            // find the nth perfect number         nthElement += 9 * outliersCount;         return nthElement;     }        // Driver Code     public static void main(String[] args)     {         System.out.println(findNth(5));     } }    // This code is contributed by // Code_Mech

Python3

 # Python3 program to find the n-th number  # with the sum of digits as 10.  import math    def findNth(n):         nthElement = 19 + (n - 1) * 9     outliersCount = int(math.log10(nthElement)) - 1        # find the nth perfect number      nthElement += 9 * outliersCount      return nthElement     # Driver Code  if __name__ == "__main__":         print(findNth(5))     # This code is contributed by Rituraj Jain

C#

 // C# program to find n-th number // with sum of digits as 10. using System;    class GFG {        public static int findNth(int n)     {         int nthElement = 19 + (n - 1) * 9;         int outliersCount = (int)Math.Log10(nthElement) - 1;            // find the nth perfect number         nthElement += 9 * outliersCount;         return nthElement;     }        // Driver Code     static public void Main()     {         Console.WriteLine(findNth(5));     } }    // This code is contributed by // Ankit Bhardwaj

Output:

55

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