Print all Integers that Aren’t Divisible by Either 2 or 3 and Lie between 1 and 50 in C++ and Python

In the previous article, we have discussed about Allocating and deallocating 2D arrays dynamically in C++. Let us learn how to Print all Integers that Aren’t Divisible by Either 2 or 3 and Lie between 1 and 50 in C++ Program.

The task is to print all the numbers from 1 to 50 which are not divisible by 2 or 3 in C++ and Python.

Sample Output:

The numbers from 1 to 50 which are not divisible by 2 and 3 are:
Number =  1
Number =  5
Number =  7
Number =  11
Number =  13
Number =  17
Number =  19
Number =  23
Number =  25
Number =  29
Number =  31
Number =  35
Number =  37
Number =  41
Number =  43
Number =  47
Number =  49

Print all Integers that Aren’t Divisible by Either 2 or 3 and Lie between 1 and 50 in C++ and Python

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There are several ways to print all the numbers from 1 to 50 which are not divisible by 2 or 3 some of them are:

Method #1: Using for loop in Python

Approach:

  • Use a for loop and a range() function which iterates from 1 to 50.
  • Then, using an if statement, determine whether the integer is not divisible by both 2 and 3.
  • If the number is not divisible by 2 and 3 then print it.
  • Exit of program

Below is the implementation:

# Use a for loop and a range() function which iterates from 1 to 50.
print("The numbers from 1 to 50 which are not divisible by 2 and 3 are:")
for numb in range(1, 51):
    # Then, using an if statement, determine whether the integer is
    # not divisible by both 2 and 3.
    if(numb % 2 != 0 and numb % 3 != 0):
        # If the number is not divisible by 2 and 3 then print it.
        print("Number = ", numb)

Output:

The numbers from 1 to 50 which are not divisible by 2 and 3 are:
Number =  1
Number =  5
Number =  7
Number =  11
Number =  13
Number =  17
Number =  19
Number =  23
Number =  25
Number =  29
Number =  31
Number =  35
Number =  37
Number =  41
Number =  43
Number =  47
Number =  49

Method #2: Using for loop in C++

Approach:

  • Use for loop in Which starts from 1 and ends at 50 using the
  • syntax : for (numb=1;numb<=50;numb++)
  • Then, using an if statement, determine whether the integer is not divisible by both 2 and 3.
  • If the number is not divisible by 2 and 3 then print it.
  • Exit of program

Below is the implementation:

#include <iostream>
using namespace std;

int main()
{
    // Use for loop in Which starts from 1 and ends at 50
    // using the syntax : for (numb=1;numb<=50;numb++)
    cout << "The numbers from 1 to 50 which are not "
            "divisible by 2 and 3 are:"
         << endl;
    for (int numb = 1; numb <= 50; numb++) {
        // Then, using an if statement determine whether the
        // integer is not divisible by both 2 and 3.

        if (numb % 2 != 0 && numb % 3 != 0) {
            // If the number is not divisible by 2 and 3
            // then print it.
            cout << "Number = " << numb << endl;
        }
    }
    return 0;
}

Output:

The numbers from 1 to 50 which are not divisible by 2 and 3 are:
Number = 1
Number = 5
Number = 7
Number = 11
Number = 13
Number = 17
Number = 19
Number = 23
Number = 25
Number = 29
Number = 31
Number = 35
Number = 37
Number = 41
Number = 43
Number = 47
Number = 49

Method #3: Using while loop in Python

Approach:

  • Take a variable  say tempo which stores the lower limit (in this case it is 1)
  • Using while loop iterate from 1 to 51 by using the condition tempo <=50
  • Then, using an if statement, determine whether the integer is not divisible by both 2 and 3.
  • If the number is not divisible by 2 and 3 then print it.
  • Increment the value of tempo by 1
  • Exit of program

Below is the implementation:

# Use a for loop and a range() function which iterates from 1 to 50.
print("The numbers from 1 to 50 which are not divisible by 2 and 3 are:")
# Take a integer variable  say tempo which stores the lower limit (in this case it is 1)
tempo = 1
# Using while loop iterate from 1 to 51 by using the condition tempo <=50
while(tempo <= 50):

    # Then, using an if statement, determine whether the integer is
    # not divisible by both 2 and 3.
    if(tempo % 2 != 0 and tempo % 3 != 0):
        # If the number is not divisible by 2 and 3 then print it.
        print('Number =', tempo)
    # Increment the value of tempo by 1
    tempo = tempo+1

Output:

The numbers from 1 to 50 which are not divisible by 2 and 3 are:
Number = 1
Number = 5
Number = 7
Number = 11
Number = 13
Number = 17
Number = 19
Number = 23
Number = 25
Number = 29
Number = 31
Number = 35
Number = 37
Number = 41
Number = 43
Number = 47
Number = 49

Method #4: Using while loop in C++

Approach:

  • Take a integer variable  say tempo which stores the lower limit (in this case it is 1)
  • Using while loop iterate from 1 to 51 by using the condition tempo <=50
  • Then, using an if statement, determine whether the integer is not divisible by both 2 and 3.
  • If the number is not divisible by 2 and 3 then print it.
  • Increment the value of tempo by 1
  • Exit of program

Below is the implementation:

#include <iostream>
using namespace std;

int main()
{
    // Use for loop in Which starts from 1 and ends at 50
    // using the syntax : for (numb=1;numb<=50;numb++)
    cout << "The numbers from 1 to 50 which are not "
            "divisible by 2 and 3 are:"
         << endl;
    // Take a integer variable  say tempo which stores the
    // lower limit (in this case it is 1)
    int tempo = 1;
    // Using while loop iterate from 1 to 51 by using the
    // condition tempo <=50
    while (tempo <= 50) {
        // Then, using an if statement determine whether the
        // integer is not divisible by both 2 and 3.

        if (tempo % 2 != 0 && tempo % 3 != 0) {
            // If the number is not divisible by 2 and 3
            // then print it.
            cout << "Number = " << tempo << endl;
        }
        // Increment the value of tempo by 1
        tempo = tempo + 1;
    }
    return 0;
}

Output:

The numbers from 1 to 50 which are not divisible by 2 and 3 are:
Number = 1
Number = 5
Number = 7
Number = 11
Number = 13
Number = 17
Number = 19
Number = 23
Number = 25
Number = 29
Number = 31
Number = 35
Number = 37
Number = 41
Number = 43
Number = 47
Number = 49

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