## Algorithm

Problem Name: beecrowd | 1070

# Six Odd Numbers

Adapted by Neilor Tonin, URI Brazil

Timelimit: 1

Read an integer value X and print the 6 consecutive odd numbers from X, a value per line, including X if it is the case.

## Input

The input will be a positive integer value.

## Output

The output will be a sequence of six odd numbers.

 Input Sample Output Sample 8 9 11 13 15 17 19

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
#include <stdio.h>
int main(){

int i, X, howManyOdd = 6;
scanf("%d", &X);
for( i = X; i  <  (X+(howManyOdd*2)) ; i+=2) {
int odd;
if(i % 2 == 0) {
odd = i + 1;
printf("%dn", odd);
} else {
odd = i;
printf("%dn", odd);
}
}
return 0;
}
``````
Copy The Code &

Input

cmd
8

Output

cmd
9 11 13 15 17 19

### #2 Code Example with C++ Programming

```Code - C++ Programming```

``````
#include <iostream>

using namespace std;

int main()
{
int x, tmp = 0;
bool ver = false;

cin >> x;

while(ver == false)
{
if(x % 2 == 1) {
cout << x << endl;
tmp++;
}

if(tmp == 6)
return 0;

x++;
}

return 0;
}
``````
Copy The Code &

Input

cmd
8

Output

cmd
9 11 13 15 17 19

### #3 Code Example with Java Programming

```Code - Java Programming```

``````
import java.util.*;

public class Main {
public static void main(String args[]) {
int i, X, howManyOdd = 6;
Scanner input = new Scanner(System.in);
X = input.nextInt();
for( i = X; i  <  (X+(howManyOdd*2)) ; i+=2) {
int odd;
if(i % 2 == 0) {
odd = i + 1;
System.out.printf("%dn", odd);
} else {
odd = i;
System.out.printf("%dn", odd);
}
}
}
}
``````
Copy The Code &

Input

cmd
8

Output

cmd
9 11 13 15 17 19

### #4 Code Example with Python Programming

```Code - Python Programming```

``````
n=int(input())
i=0
while(i<6):
if(n%2!=0):
print(n)
i=i+1
n=n+1
``````
Copy The Code &

Input

cmd
8

Output

cmd
9 11 13 15 17 19