# Numbers less than N that are perfect cubes and the sum of their digits reduced to a single digit is 1

Given a number n, the task is to print all the numbers less than or equal to n which are perfect cubes as well as the eventual sum of their digits is 1.

Examples:

Input: n = 100
Output: 1 64
64 = 6 + 4 = 10 = 1 + 0 = 1
Input: n = 1000
Output: 1 64 343 1000

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

Approach: For every perfect cube less than or equal to n keep on calculating the sum of its digits until the number is reduced to a single digit ( O(1) approach here ), if this digit is 1 then print the perfect cube else skip to the next perfect cube below n until all the perfect cubes have been considered.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Functin that returns true if the eventual ` `// digit sum of number nm is 1 ` `bool` `isDigitSumOne(``int` `nm) ` `{ ` `   ``//if reminder will 1  ` `   ``//then eventual sum is 1 ` `    ``if` `(nm % 9 == 1) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Function to print the required numbers ` `// less than n ` `void` `printValidNums(``int` `n) ` `{ ` `    ``int` `cbrt_n = (``int``)cbrt(n); ` `    ``for` `(``int` `i = 1; i <= cbrt_n; i++) { ` `        ``int` `cube = ``pow``(i, 3); ` ` `  `        ``// If it is the required perfect cube ` `        ``if` `(cube >= 1 && cube <= n && isDigitSumOne(cube)) ` `            ``cout << cube << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 1000; ` `    ``printValidNums(n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``// Functin that returns true if the eventual ` `    ``// digit sum of number nm is 1 ` `    ``static` `boolean` `isDigitSumOne(``int` `nm) ` `    ``{ ` ` `  `      ``//if reminder will 1  ` `      ``//then eventual sum is 1 ` `      ``if` `(nm % ``9` `== ``1``) ` `        ``return` `true``; ` `      ``else` `        ``return` `false``; ` `    ``} ` ` `  `    ``// Function to print the required numbers ` `    ``// less than n ` `    ``static` `void` `printValidNums(``int` `n) ` `    ``{ ` `        ``int` `cbrt_n = (``int``)Math.cbrt(n); ` `        ``for` `(``int` `i = ``1``; i <= cbrt_n; i++) { ` `            ``int` `cube = (``int``)Math.pow(i, ``3``); ` ` `  `            ``// If it is the required perfect cube ` `            ``if` `(cube >= ``1` `&& cube <= n && isDigitSumOne(cube)) ` `                ``System.out.print(cube + ``" "``); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``1000``; ` `        ``printValidNums(n); ` `    ``} ` `} `

## Python

 `# Python3 implementation of the approach ` `import` `math ` ` `  `# Functin that returns true if the eventual  ` `# digit sum of number nm is 1 ` `def` `isDigitSumOne(nm) : ` `   ``#if reminder will 1  ` `   ``#then eventual sum is 1 ` `    ``if``(nm ``%` `9` `=``=` `1``): ` `        ``return` `True` `    ``else``: ` `        ``return` `False` ` `  `# Function to print the required numbers ` `# less than n ` `def` `printValidNums(n): ` `    ``cbrt_n ``=` `math.ceil(n``*``*``(``1.``/``3.``)) ` `    ``for` `i ``in` `range``(``1``, cbrt_n ``+` `1``): ` `        ``cube ``=` `i ``*` `i ``*` `i ` `        ``if` `(cube >``=` `1` `and` `cube <``=` `n ``and` `isDigitSumOne(cube)): ` `            ``print``(cube, end ``=` `" "``) ` `             `  ` `  `# Driver code  ` `n ``=` `1000` `printValidNums(n) `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// Functin that returns true if the ` `    ``// eventual digit sum of number nm is 1 ` `    ``static` `bool` `isDigitSumOne(``int` `nm) ` `    ``{ ` ` `  `     ``//if reminder will 1  ` `     ``//then eventual sum is 1 ` `      ``if` `(nm % 9 == 1) ` `         ``return` `true``; ` `      ``else` `         ``return` `false``; ` `    ``} ` ` `  `    ``// Function to print the required  ` `    ``// numbers less than n ` `    ``static` `void` `printValidNums(``int` `n) ` `    ``{ ` `        ``int` `cbrt_n = (``int``)Math.Ceiling(Math.Pow(n,  ` `                                      ``(``double``) 1 / 3)); ` `        ``for` `(``int` `i = 1; i <= cbrt_n; i++)  ` `        ``{ ` `            ``int` `cube = (``int``)Math.Pow(i, 3); ` ` `  `            ``// If it is the required perfect cube ` `            ``if` `(cube >= 1 && cube <= n &&  ` `                             ``isDigitSumOne(cube)) ` `                ``Console.Write(cube + ``" "``); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``int` `n = 1000; ` `        ``printValidNums(n); ` `    ``} ` `} ` ` `  `// This code is contributed by akt_mit  `

## PHP

 `= 1 && ``\$cube` `<= ``\$n` `&&  ` `                    ``isDigitSumOne(``\$cube``))  ` `            ``echo` `\$cube``, ``" "``;  ` `    ``}  ` `}  ` ` `  `// Driver code  ` `\$n` `= 1000;  ` `printValidNums(``\$n``);  ` ` `  `// This code is contributed by Ryuga ` `?> `

Output:

```1 64 343 1000
```

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