Help w/ divisors by 4

For the integers n in the range of 60 to 100, print each on its own row as well as the divisors of n that are evenly divisible by 4.

Code:

---

#include <stdio.h>

int main (void)
{

    int a,
        sum = 0;

        for( a = 60; a <= 100; a++)
        {
            sum += a;
                printf("%d:\n", a);
        }

}

---

With this I have it printing:
60:
61:
62:
....
....
100:


I am looking to print something like this:
60: 4 12 20 60
...
....
.....
100: 4 20 100

What would I do to print that out? Use a mod?

yeah, % is the modulus operator. So, if n % 4 == 0 then n is divisible by 4

Note all divisors of a given n that are divisible by 4 will be multiples of 4 also.
Based on this you can insert a nested loop that calculates that for a given n.

Quote:

---

Originally Posted by BB89

For the integers n in the range of 60 to 100, print each on its own row as well as the divisors of n that are evenly divisible by 4.

...

I am looking to print something like this:
60: 4 12 20 60
...
....
.....
100: 4 20 100

What would I do to print that out? Use a mod?

---

Yes.
for i from 60 to 100
Determine whether the number(the i in your for loop) is divisible by the divisor ( = 4) .
if (yes)
then print divisor and then
Go on checking the multiples of your divisor whether they are divisible by i .If yes print it.


Sample run.

Code:

---

60 : 4 12 20 60
64 : 4 8 16 32 64
68 : 4 68
72 : 4 8 12 24 36 72
76 : 4 76
80 : 4 8 16 20 40 80
84 : 4 12 28 84
88 : 4 8 44 88
92 : 4 92
96 : 4 8 12 16 24 32 48 96
100 : 4 20 100

---

I've got this so far.

60:
64:
68:
72:
76:
80:
84:
88:
92:
96:
100:

Code:

---

#include <stdio.h>

int main (void)
{
    int a,
        sum = 0;

        for( a = 60; a <= 100; a++)
            if( a % 4 == 0 )
            {
                sum += a;
                    printf("%d:\n", a);
            }
}

---

still having trouble printing the rest of it.

Code:

---

#include <stdio.h>

int main (void)
{
    int a,
        sum = 0;

        for( a = 60; a <= 100; a++)
            if( a % 4 == 0 )
            {
                /* What for  : sum += a;*/
                    printf("%d:\n", a);
            /*Now check for multiples of 4
              Is a % 8 == 0  if yes print 8
              Is a % 12 == 0  if yes print 12
              Is a % 16 == 0 if yes print 16
              Use a loop. which runs from 8 to a and checks the above
          */
              }
}

---

Quote:

---

Originally Posted by BB89

For the integers n in the range of 60 to 100, print each on its own row as well as the divisors of n that are evenly divisible by 4.

Code:

---

#include <stdio.h>

int main (void)
{

    int a,
        sum = 0;(why select sum=0 initially)

        for( a = 60; a <= 100; a++)
        {
            sum += a;-------- meaning of this
                printf("%d:\n", a);
        }

}

---

With this I have it printing:
60:
61:
62:
....
....
100:


I am looking to print something like this:
60: 4 12 20 60
...
....
.....
100: 4 20 100

What would I do to print that out? Use a mod?

---

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tyutyutyutyutyutyuy

What you are doing is checking if the numbers between 60 and 100 are divisible by 4. That is not what the question is asking. You need to:

find the divisors of each numbers between 60 and 100.
Test if the numbers you found in the previous step are divisible by 4

In pseudocode:

Code:

---


for a in [60,100] do
      for i in [0,a] do
          if (i divides a AND 4 divides i) then
                print i
          end if
      end for
end for

---