CODEWITHSUNDEEP
c#-3.0
harpreet
harpreet

Reputation: 19

C# for loop to print first 20 positive numbers divisible by 7?

This is Code I reached. What I don't understand is how do I put a condition to display the first 20 numbers where I wrote the condition for i to be less than 20. I know that my code is completely wrong.

 for(int i=1; i<=20; i++)
        {
            if(i%7==0)
            {
                Console.WriteLine(i);
            }
        }

Upvotes: 0

Views: 1311

Answers (4)

Zahidul
Zahidul

Reputation: 76

You can try bellow code that will give you first 20 numbers that will covered the condition i%7==0 ........

  public static void Main(string[] args)
    {
       int i=0, count = 0;
       while(count < 20)
        {
            if (i % 7 == 0)
            {
                Console.WriteLine("Position {0} number is = {1}", count+1, i,"\n");
                count++;
            }
            i++;
        }
        Console.ReadKey();
    }

Upvotes: 0

Ousmane D.
Ousmane D.

Reputation: 56453

You're close. use a counter variable:

int counter = 0; // counter variable
for(int i=1; ; i++) // removed condition 
{
       if (counter > 20) break; // time to stop the iteration
       if(i%7==0)
       {
            counter++;
            Console.WriteLine(i);
       }
}

This can be improved to:

for(int i = 7, counter = 0; counter <= 20; i += 7) 
{                     
       Console.WriteLine(i);
       counter++;
}

Upvotes: 2

GrahamS
GrahamS

Reputation: 10350

Can’t you just go up in sevens?

for (int multiple = 7, int count = 0; count < 20; multiple += 7, count++)
{
    Console.WriteLine(multiple);
}

Upvotes: 1

kvantour
kvantour

Reputation: 26501

The first 20 integers that are divisible by 7 are easily written as 7,2*7,3*7,4*7,...,20*7. This in your loop you can do:

for(int i = 1; i<=20; i++) {
   Console.WriteLine(7*i);
}

Upvotes: 1

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