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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!!

tonja1196
Newbie Poster
9 posts since Oct 2004
Reputation Points: 10
Solved Threads: 0
 

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 afor 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 expressiontrialDivisor <= 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

Stack Overflow
Junior Poster
193 posts since Sep 2004
Reputation Points: 26
Solved Threads: 4
 

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 afor 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 expressiontrialDivisor <= 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

Thank You!!

tonja1196
Newbie Poster
9 posts since Oct 2004
Reputation Points: 10
Solved Threads: 0
 

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)

tonja1196
Newbie Poster
9 posts since Oct 2004
Reputation Points: 10
Solved Threads: 0
 

>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.

Narue
Bad Cop
Administrator
15,460 posts since Sep 2004
Reputation Points: 6,464
Solved Threads: 1,401
 

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
#include
using namespace std;
#include

int main() {
int num = 20,divisor,divisorSum,total=0,num2=490,count;


cout<num)
{ cout<< setw(4)<<"\t"<

tonja1196
Newbie Poster
9 posts since Oct 2004
Reputation Points: 10
Solved Threads: 0
 

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