This is clean! Awesome!

What have you imported for this code?

public class Solution {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        for (int i = 1; i <= 10 && input.hasNextInt(); i++) {
            int n = input.nextInt();
            System.out.printf("%d x %d = %d\n", n, i, (n*i));
        }
    }
}

Mine wont work^^?

it should be \n instead of %n

Will you explain me: System.out.printf("%d x %d = %d%n", N, i, N*i);

can you please explain sir why did you use "printf" function instead of "println" function and kindly could you tell me its purpose?

Exactly Correct.Thanks for uploading solution

i m facing this issue, can u pls help me?

error: no suitable method found for println(String,int,int,int)

can u explain this!!without getting a value how this get printed and ya also you are using printf here

I Need a Help Regarding N limit N>=2&&N<=20 No one Gave The range of N so for this challenge if n=30 the following answer will not work.

can you explain this : %d x %d

will this pass the test case with N=45

how i print x ??? shift +8=*.. but its wrong????why?

private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
    int N = scanner.nextInt();
    scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

    for(int i=1;i<=10;i++)
    {
                    // Can use \n instead of %n, both works fine
        System.out.printf("%d x %d = %d\n",N,i,N*i);
    }
    scanner.close();
}

Why aren't you setting up the constraint for N, i.e. 2<=N<=20?

in the for loop instead of using only i y do u use integer type i. u cn show it in above.

does it take the value of N throgh user

what is the diffrence between these two lines

System.out.println("%d x %d = %d ",N,i,N*i); &
System.out.printf("%d x %d = %d%n", N, i, N*i);

Why can't we use println here instead of printf, why is it showing an error

But,where is the cross symbol in keyboard?

why println is not working for this code

but where did you add the contraints for N?

test-0 is passed but test-1 and test-2 is failled why

why %n ?

Why does this code not print a nextline in the 10th iteration?

So you are using System.out.println(); in C ??

;)

this is exactly how i solved it, but need to keep using printf, i think, since everyone uses it here.

can't understand why mine gives the error. public static void main(String[] args) { int N = scanner.nextInt(); scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

for(int i=1; i<=10;i++){ int z = N*i; } System.out.println(N + " x " + i + " = " + z); scanner.close(); }

what is the use of scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");

remove this"scanner.skip......" and simply write your own code.