## paul7048's Profile User Rating:

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1. #### In Topic: Making a program that counts perfect, deficient, and abundant numbers?

Posted 14 Sep 2011

Thank you Adak, it runs through now.
2. #### In Topic: Making a program that counts perfect, deficient, and abundant numbers?

Posted 14 Sep 2011

I've been reading a lot of topics and posts, and Dream.In.Code. seems like a great supporting environment. Nice to meet you Adak. Just being curious is your name from the middle east region?
3. #### In Topic: Making a program that counts perfect, deficient, and abundant numbers?

Posted 14 Sep 2011

For my C++ class, I have to make a program that displays all numbers that are perfect, deficient, and abundant below 5,000. A number is called perfect if the sum of its divisors is equal to the original number, a number is called deficient if the sum of its divisors is less than the original number, and a number is called abundant if the sum of its divisors is more than the original number.

I've made a "pre-program" that's going to I'm using to illustrate what the finished program will be doing, but I seem to be getting an error at the "else (counter<sum)" saying that there is no semicolon?

I set the global constant to 15, and if the sum is equal to 6, then it will say that number is perfect, if the sum is greater than 6, it will say it's abundant, and if less than 6 then it's deficient.

Here is the code I have so far:

```#include <iostream>
using namespace std;
//global constant
const int maxvalue = 15;
int sumofDivisor(int);
int main()
{
int counter, sum;
//for loop to max value
sum=sumofDivisor(6);
for(counter=1; counter<=maxvalue; counter++)

if(counter==sum)

{
cout << " this number is perfect";
}
else if (counter>sum)
{
cout << "these are abundant";
}
else (counter<sum)
{
cout << "they're deficient";
}

return 0;
}

int sumofDivisor(int a)
{

return a;
}
```

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