A quick and easy problem (not for me though)

[MySecretName]

Hello,

I have a problem in C:

Print all numbers in this form: a23a, where a is a digit, which are divisible by 6;

I know the divisible part(if(x%6==0), etc.) but i'm having problems with the a23a;

The problem must be solve in the easiest way possible.
Thank you!

[Adak]

printf() gives you the complete control you need for this.

print the divisible int, then 23, then the divisible int variable again. Print four numbers, with no spaces between them.

Post up the code and we can see where the problem is. Right now, I can't understand what the stumbling block is.

[JonathanS]

Don't quite understand what you want.

you want to print all the combinations of a23a and a is divisible by 6?

so for positive numbers output should be:

6236
122312
182318
so on...

If so it will be an infinite loop and go on forever. Is there a max you put to a?

[MySecretName]

Don't quite understand what you want.

you want to print all the combinations of a23a and a is divisible by 6?

so for positive numbers output should be:

6236
122312
182318
so on...

If so it will be an infinite loop and go on forever. Is there a max you put to a?

Not an ifinite loop, just the numbers that have 23 in the middle , for example:

1231
1232
3235
9237
...

All the combinations of 4 digits with 23 in the middle.

@Adak: Thanks, I'm trying to figure it out right now.

LATER EDIT:

@Adak: using printf does not alow me to anything with the output number, for example:

int i;

for(i=0; i<10; i++)
{
    printf("%d23%d\n", i,i);
}

...

I need some way to store the a23a, so i can put it inside an if condition and see if my a23a is divided by 6. If it's divided by 6 i will printf it, else do nothing.

[JonathanS]

I don't get the reasoning for the output. What is supposed to be divisible by 6?

1231 is not divisible by 6. neither is a (which is one in this case).

is a+2+3 divisible by 6 you mean?

[Adak]

So the first one OR two digits (as in 1231), must be divisible by 6, and the last digit can be any digit at all? (3235)???.

Problem is, 92 is not divisible by 6, (90 is and so is 96 though), and neither is 9 of course. You WILL need a top end stopping point for the loop, even for preliminary testing. Maybe 50,000 to start?

[MySecretName]

I'll give you a PHP example to be more clear:

<?php

for($i=0; $i<10; $i++)
{
    $mynumber = $i."23".$i;

    if($mynumber%6 == 0)
        echo $mynumber."<br />";
}

?>

OUTPUT:

2232
8238

The C program should check just this combinations(a23a -> a is the same digit):

1231
2232
3233
4234
5235
6236
7237
8238
9239

[Adak]

OK, this part, I understand - very simple:

The C program should check just this combinations(a23a -> a is the same digit):

1231
2232
3233
4234
5235
6236
7237
8238
9239

But 2 mod six is not 0, so the mynumber % 6==0, part, is confusing me. As in: 2232

You could print out "23" as a string:

printf(%d%s%d \n", mynumber, string23, mynumber);

or you could make 23 into a digit:

int twenty3 =23;

printf("%d%d%d", mynumber, twenty3, mynumber);

In either case, you could use a for loop, very much like your PHP example.

Do you want to generate these combinations, or simply check them that they have this feature?

[MySecretName]

I need to generate them, and every time i generate a number i need to check if the number is divided by 6.

#include <stdio.h>
#include <stdlib.h>

int i,a;

int main(int argc, char *argv[])
{
    a = 23;
    
    for(i=0; i<10; i++)
    {
             printf("%d%d%d\n", i, a, i);
    }
  
  system("PAUSE");    
  return 0;
}

How does this code is supposed to look like, where the if should be in order to output just 2232,8238?

[CommonTater]

Hello,

I have a problem in C:

Print all numbers in this form: a23a, where a is a digit, which are divisible by 6;

I know the divisible part(if(x%6==0), etc.) but i'm having problems with the a23a;

The problem must be solve in the easiest way possible.
Thank you!

Ok if a is a single digit... there is only 1 possibility ... 6236 ... there all done.

[MySecretName]

How is 6236 divided by 6?

Google

[CommonTater]

How is 6236 divided by 6?

Google

Ok... so you want to divide the whole number, not just the two "tacked on" digits...

Carry on....

[MySecretName]

Ok, thanks anyway for your help.
Close this thread please, you are to advanced to solve this, i understand.

[quzah]

I'm still confused about what you are trying to do here. Why are you singling out "23"?

1230 is divisible by 6.


Quzah.

[CommonTater]

Ok, thanks anyway for your help.
Close this thread please, you are to advanced to solve this, i understand.

Naaa... just making a little joke then steppin' aside so the others can take their turn...

I don't know who is giving out these bizarre assignments but I've never seen anything like this one come up in real-world programming.