Thread: Simple C program displaying all numbers divisible by 6

  1. #1
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    Smile Simple C program displaying all numbers divisible by 6

    Hey!!
    Was just wondering if its possible to do a simple C program displays all the numbers divisible by 6 in the range of 1 to 40 in rows of 3 columns...its kinda complicated to me..can anybody help me with a hint?? or explian??

    thankies

  2. #2
    C++ Witch laserlight's Avatar
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    First, write a program that finds and "displays all the numbers divisible by 6 in the range of 1 to 40". Then think about the displaying in rows of 3 columns problem.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  3. #3
    Frequently Quite Prolix dwks's Avatar
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    The % operator is useful for both parts of the problem as laserlight has outlined it . . . .
    dwk

    Seek and ye shall find. quaere et invenies.

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  4. #4
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    Thanks for who tried to help..
    I did this so far...

    Code:
    #include <stdio.h>
    void main()
    {
    int x=0;
    for(x=3;x<10; x+=3)
    printf("&#37;i\t" ,x);
    printf("\n") ;
    for(x=12;x<19; x+=3)
    printf("%i\t" ,x);
    printf("\n") ;
    for(x=21;x<28; x+=3)
    printf("%i\t" ,x);
    printf("\n") ;
    for(x=30;x<37; x+=3)
    printf("%i\t" ,x);
    printf("\n") ;
    for(x=39;x<40; x+=3)
    printf("%i", x);
    }

    But,I need the program to calculate it..can any body figure my mistake which I on't get....
    thankies again

  5. #5
    Deathray Engineer MacGyver's Avatar
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    Well, that's one way to overly complicate something simple.

  6. #6
    Frequently Quite Prolix dwks's Avatar
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    Like I said, the &#37; operator is useful in two cases here. You can use it to check if a number is divisible by six.
    Code:
    if(n % 6 == 0) { /* it's divisible by 6 */ }
    You can also use it to space out your columns. Think of it this way: when the number of numbers you have printed is a multiple of 3, print a newline.
    dwk

    Seek and ye shall find. quaere et invenies.

    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
    "Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
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  7. #7
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    Thanks! You helped me in the first part : D

  8. #8
    Frequently Quite Prolix dwks's Avatar
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    Yeah, because I gave you the exact code required.
    dwk

    Seek and ye shall find. quaere et invenies.

    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
    "Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
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  9. #9
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    I think I have done the code...but,with some goofs lol :P

    Code:
    #include <stdio.h>
    int main()
    {
    int x,count=0;
    for(count>=1 && count>=40;)
    {
    if(x &#37;3==0)
    printf("%i/n",x);
    }
    
    }

  10. #10
    and the hat of int overfl Salem's Avatar
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    If x is a number which is divisible by 6, then doing x = x + 6 also results in a number divisible by 6.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  11. #11
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    Code:
    #include <stdio.h>
    
    int main(void)
    {
    	int i;
    	int j;
    
    	j = 0;
    	for (i = 1; i <= 40; i++)
    	{
    		if (j < 3)
    		{
    			if (i % 6 == 0)
    			{
    				printf("\t%d\t", i);
    				j++;
    			}
    		}
    		else
    		{       if ( j == 3)
    			{
    				j = 0;
    				printf("\n");
    			}
    		}
    	}
    
    	return (0);
    }

  12. #12
    C++ Witch laserlight's Avatar
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    broli86, read our homework policy.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  13. #13
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    Ok , I appologize.

  14. #14
    Frequently Quite Prolix dwks's Avatar
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    Your solution skips over numbers where j happens to be 3.

    And anyway, you're just overcomplicating as well. You can do it much more simply. Since we're giving away solutions . . . .
    Code:
    #include <stdio.h>
    
    int main() {
        int num, count = 0;
    
        for(num = 1; num <= 40; num ++) {
            if(num &#37; 6 == 0) {
                printf("%3d", num);
    
                if(++count % 3 == 0) putchar('\n');
            }
    
        }
    
        return 0;
    }
    (Most of the time I wouldn't give away something like this, but the other program works too and has been on the internet for a long time already -- I'm sure the OP has seen it.)
    dwk

    Seek and ye shall find. quaere et invenies.

    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
    "Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
    "The only real mistake is the one from which we learn nothing." -- John Powell


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  15. #15
    Deathray Engineer MacGyver's Avatar
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    I think we're skipping the most obvious solution. Salem hinted at it.

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