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:**

### 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:**

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