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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