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Not Yet Answered # Loop Issues

I have a lab to create a loop to determine the positive divisors of an interger and then determine if the interger is a perfect,deficient, etc. I am having trouble determining how to write the statement to determine the divisors of the numbers. I believe the % operator is used. Just not sure. Help!!

Stack Overflow 8 Discussion Starter tonja1196 Discussion Starter tonja1196 Narue 5,707 Discussion Starter tonja1196 Hi I'm having a problem implementing a mini shopping cart drop down in the header to show the user all the products they have in their shopping cart. It seems the only solution for this is Ajax, and I've looked all over and can't find anything that I could possibly ...

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Greetings,

Well, you're right on. Finding the divisor of a positive number can be easily found using the modulus operator, %. The modulus operator works on integers (and integer expressions) and yields the remainder when the first operand is divided by the second. In C++, the modulus operator is a percent sign, %. The syntax is exactly the same as for other operators:

```
int quotient = 7 / 3;
int remainder = 7 % 3;
```

The first operator, integer division, yields 2. The second operator yields 1. Thus, 7 divided by 3 is 2 with 1 left over.

The modulus operator turns out to be surprisingly useful. For example, you can check whether one number is divisible by another: if x % y is zero, then x is divisible by y.

Now lets use this same principal by finding the divisor of 12. We know 1, 2, 3, 4, 6, and 12 are divisible by twelve. Maybe this can help us figure out how to determine the divisors using a for loop.

```
#include <stdio.h>
int main() {
int num = 12;
printf("The divisors of 12 are ");
for (trialDivisor = 1; trialDivisor <= num / 2; ++trialDivisor)
if (num % trialDivisor == 0) {
divisorSum += trialDivisor; // in other words: divisor_sum = divisor_sum + trial_divisor;
printf("%d ", trialDivisor);
}
return 0;
}
```

This example is quite simple. You might ask yourself why we call the expression **trialDivisor <= num / 2;**. Well, I believe that is because 2 will always be a divisor of a positive number. It's usually **dividend / divisor**, and a divisor of 2 will never fail. Plus, we know that 24 is divisible into 24, though divisors only start at '/ 2' and down.

24 is divisbile by 12, 8, 6, etc...

36 is divisible by 18, 12, etc...

Hope this helps,

- Stack Overflow

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Greetings,

Well, you're right on. Finding the divisor of a positive number can be easily found using the modulus operator, %. The modulus operator works on integers (and integer expressions) and yields the remainder when the first operand is divided by the second. In C++, the modulus operator is a percent sign, %. The syntax is exactly the same as for other operators:

`int quotient = 7 / 3; int remainder = 7 % 3;`

The first operator, integer division, yields 2. The second operator yields 1. Thus, 7 divided by 3 is 2 with 1 left over.

The modulus operator turns out to be surprisingly useful. For example, you can check whether one number is divisible by another: if x % y is zero, then x is divisible by y.

Now lets use this same principal by finding the divisor of 12. We know 1, 2, 3, 4, 6, and 12 are divisible by twelve. Maybe this can help us figure out how to determine the divisors using a for loop.

`#include <stdio.h> int main() { int num = 12; printf("The divisors of 12 are "); for (trialDivisor = 1; trialDivisor <= num / 2; ++trialDivisor) if (num % trialDivisor == 0) { divisorSum += trialDivisor; // in other words: divisor_sum = divisor_sum + trial_divisor; printf("%d ", trialDivisor); } return 0; }`

This example is quite simple. You might ask yourself why we call the expression

trialDivisor <= num / 2;. Well, I believe that is because 2 will always be a divisor of a positive number. It's usuallydividend / divisor, and a divisor of 2 will never fail. Plus, we know that 24 is divisible into 24, though divisors only start at '/ 2' and down.24 is divisbile by 12, 8, 6, etc...

36 is divisible by 18, 12, etc...

Hope this helps,

- Stack Overflow

Thank You!!

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Thank you so much for the help. If the divisors of 12 are 1,2,3,4,6 what do I need to do to show the total of the divisors?? : (1+2+3+4+6=16)

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>what do I need to do to show the total of the divisors?

Keep a running total. In StackOverflow's example, look at divisorSum that he neglected to declare. That's the general idea.

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See my attachement below: I believe one one braces is in the wrong location. I am trying to loop a range of #20-30 and #490-500 and have each number printed and stating is the number is deficient,perfect or abundant. The ouput I am tryint to achieve should be in this format. I hope this is making sense.

```
Number Classification
20 deficient*(*I am guessing the outcome)
21 perfect
"" ""
500 abundant
#include <stdio.h>
#include <iostream>
using namespace std;
#include<iomanip>
int main() {
int num = 20,divisor,divisorSum,total=0,num2=490,count;
cout<<setw(2)<< "\n Number Classification \n";
cout <<"___________________________________" <<"\n" ;
while (num<=30,num2<=500)
{ cout<< num<<num2;
num++,num2++;
for (divisor = 1; divisor <= num / 2; ++divisor)
if (num % divisor == 0)
{
total=total+divisor;
}
if (total>num)
{ cout<< setw(4)<<"\t"<<num<< "\tAbundant \n";
cout<< setw(4)<<"\t"<<num2<< "\tAbundant \n" ;
}
else if (total<num)
{ cout<< setw(4)<<"\t"<<num<<"\tDeficient \n";
cout<< setw(4)<<"\t"<<num2<<"\tDeficient \n";
}
else if (total=num)
{ cout<< setw(4)<<"\t"<<num<<"\tPerfect \n";
cout<< setw(4)<<"\t"<<num2<<"\tPerfect \n";
}
}
system ("pause");
return 0;
}
```

*Edited 3 Years Ago by Dani*: Formatting fixed

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