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 "````, 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 `

/div>

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 9, 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 "````, 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++

 `// Java 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
```