# 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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