# Sum of all the prime numbers with the count of digits ≤ D

Given an integer D, the task is to find the sum of all the prime numbers whose count of digits is less than or equal to D.

Examples:

Input: D = 2
Output: 1060
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,
47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97 are
the prime numbers having digits less than or equal
to 2 and the sum of these prime numbers is 1060.

Input: D = 3
Output: 76127

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

Approach: Generate all prime numbers using Sieve of Eratosthenes upto maximum D-digit number then find the sum of all the prime numbers in the same range.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function for Sieve of Eratosthenes ` `void` `sieve(``bool` `prime[], ``int` `n) ` `{ ` `    ``prime[0] = ``false``; ` `    ``prime[1] = ``false``; ` `    ``for` `(``int` `p = 2; p * p <= n; p++) { ` `        ``if` `(prime[p] == ``true``) { ` `            ``for` `(``int` `i = p * p; i <= n; i += p) ` `                ``prime[i] = ``false``; ` `        ``} ` `    ``} ` `} ` ` `  `// Function to return the sum of ` `// the required prime numbers ` `int` `sumPrime(``int` `d) ` `{ ` ` `  `    ``// Maximum number of d-digits ` `    ``int` `maxVal = ``pow``(10, d) - 1; ` ` `  `    ``// Sieve of Eratosthenes ` `    ``bool` `prime[maxVal + 1]; ` `    ``memset``(prime, ``true``, ``sizeof``(prime)); ` `    ``sieve(prime, maxVal); ` ` `  `    ``// To store the required sum ` `    ``int` `sum = 0; ` ` `  `    ``for` `(``int` `i = 2; i <= maxVal; i++) { ` ` `  `        ``// If current element is prime ` `        ``if` `(prime[i]) { ` `            ``sum += i; ` `        ``} ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `d = 3; ` ` `  `    ``cout << sumPrime(d); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` `     `  `    ``// Function for Sieve of Eratosthenes  ` `    ``static` `void` `sieve(``boolean` `[]prime, ``int` `n)  ` `    ``{  ` `        ``prime[``0``] = ``false``;  ` `        ``prime[``1``] = ``false``;  ` `        ``for` `(``int` `p = ``2``; p * p <= n; p++) ` `        ``{  ` `            ``if` `(prime[p] == ``true``)  ` `            ``{  ` `                ``for` `(``int` `i = p * p; i <= n; i += p)  ` `                    ``prime[i] = ``false``;  ` `            ``}  ` `        ``}  ` `    ``}  ` `     `  `    ``// Function to return the sum of  ` `    ``// the required prime numbers  ` `    ``static` `int` `sumPrime(``int` `d)  ` `    ``{  ` `        ``int` `i; ` `        ``// Maximum number of d-digits  ` `        ``int` `maxVal = (``int``)Math.pow(``10``, d) - ``1``;  ` `     `  `        ``// Sieve of Eratosthenes  ` `        ``boolean` `prime[] = ``new` `boolean``[maxVal + ``1``]; ` `         `  `        ``for``(i = ``0``; i < maxVal + ``1``; i++) ` `            ``prime[i] = ``true``; ` `             `  `        ``sieve(prime, maxVal);  ` `     `  `        ``// To store the required sum  ` `        ``int` `sum = ``0``;  ` `     `  `        ``for` `(i = ``2``; i <= maxVal; i++)  ` `        ``{  ` `     `  `            ``// If current element is prime  ` `            ``if` `(prime[i])  ` `            ``{  ` `                ``sum += i;  ` `            ``}  ` `        ``}  ` `        ``return` `sum;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `d = ``3``;  ` `     `  `        ``System.out.println(sumPrime(d));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach  ` `from` `math ``import` `sqrt ` ` `  `# Function for Sieve of Eratosthenes  ` `def` `sieve(prime, n) : ` ` `  `    ``prime[``0``] ``=` `False``;  ` `    ``prime[``1``] ``=` `False``;  ` `    ``for` `p ``in` `range``(``2``, ``int``(sqrt(n)) ``+` `1``) : ` `        ``if` `(prime[p] ``=``=` `True``) :  ` `            ``for` `i ``in` `range``( p ``*` `p, n ``+` `1``, p) : ` `                ``prime[i] ``=` `False``;  ` ` `  `# Function to return the sum of  ` `# the required prime numbers  ` `def` `sumPrime(d) : ` ` `  `    ``# Maximum number of d-digits  ` `    ``maxVal ``=` `(``10` `*``*` `d) ``-` `1``;  ` ` `  `    ``# Sieve of Eratosthenes  ` `    ``prime ``=` `[``True``] ``*` `(maxVal ``+` `1``);  ` `    ``sieve(prime, maxVal);  ` ` `  `    ``# To store the required sum  ` `    ``sum` `=` `0``;  ` ` `  `    ``for` `i ``in` `range``(``2``, maxVal ``+` `1``) :  ` ` `  `        ``# If current element is prime  ` `        ``if` `(prime[i]) : ` `            ``sum` `+``=` `i;  ` ` `  `    ``return` `sum``;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``d ``=` `3``;  ` ` `  `    ``print``(sumPrime(d));  ` ` `  `# This code is contributed by kanugargng `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` `     `  `class` `GFG  ` `{ ` `     `  `    ``// Function for Sieve of Eratosthenes  ` `    ``static` `void` `sieve(Boolean []prime, ``int` `n)  ` `    ``{  ` `        ``prime[0] = ``false``;  ` `        ``prime[1] = ``false``;  ` `        ``for` `(``int` `p = 2; p * p <= n; p++) ` `        ``{  ` `            ``if` `(prime[p] == ``true``)  ` `            ``{  ` `                ``for` `(``int` `i = p * p;  ` `                         ``i <= n; i += p)  ` `                    ``prime[i] = ``false``;  ` `            ``}  ` `        ``}  ` `    ``}  ` `     `  `    ``// Function to return the sum of  ` `    ``// the required prime numbers  ` `    ``static` `int` `sumPrime(``int` `d)  ` `    ``{  ` `        ``int` `i; ` `        ``// Maximum number of d-digits  ` `        ``int` `maxVal = (``int``)Math.Pow(10, d) - 1;  ` `     `  `        ``// Sieve of Eratosthenes  ` `        ``Boolean []prime = ``new` `Boolean[maxVal + 1]; ` `         `  `        ``for``(i = 0; i < maxVal + 1; i++) ` `            ``prime[i] = ``true``; ` `             `  `        ``sieve(prime, maxVal);  ` `     `  `        ``// To store the required sum  ` `        ``int` `sum = 0;  ` `     `  `        ``for` `(i = 2; i <= maxVal; i++)  ` `        ``{  ` `     `  `            ``// If current element is prime  ` `            ``if` `(prime[i])  ` `            ``{  ` `                ``sum += i;  ` `            ``}  ` `        ``}  ` `        ``return` `sum;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main (String[] args) ` `    ``{  ` `        ``int` `d = 3;  ` `     `  `        ``Console.WriteLine(sumPrime(d));  ` `    ``}  ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```76127
```

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