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# Finding Divisors

In the code below it is not registering float values - for example if it divides 4 by 3 it is registering 1 as an answer (if you enter 4 when it loops up to 3 it doesn't record the float value).

I want to store the float value so I can see if it is a divisor or not.

I want to store the float value so I can see if it is a divisor or not.

```
#include <iostream>
#include <cmath>
bool is_integer(float number) {
return std::floor(number) == number;
}
int divisors(int number) {
int i = 1;
float result;
for (i = 1; i<= number; i++) {
result = number / i;
std::cout << i;
std::cout << result;
std::cout << "\n";
}
}
int main()
{
int counter = 1;
int number;
int number_divisors;
std::cout << "Enter a number to find the divisors";
std::cin >> number;
number_divisors = divisors(number);
return 0;
}
```

Integer math will do this job faster than floating point. Only numbers up to the square root of the number under test must be examined. Example pseudocode below ...

```
for k = 2 to sqrt(testnumber)
if [testnumber mod k) == 0] {
print (k, "is a factor of", testnumber)
print (testnumber/k, "is a factor of", testnumber)
}
next k
```

result = number / i;you could make it 2 statements as suggested by phoffric or do

`result = ((float)number )/ i;`

this would 'cast' the nominator to float and the division result is float as well.

Sara

`number_divisors = divisors(number);`

you seem to be expecting divisors() to return something but I don't see a return statement in the function.

```
float result;
for (i = 1; i<= number; i++) {
result = number / i;
std::cout << i;
std::cout << result;
std::cout << "\n";
}
}
```

also if you are trying to count divisors the logic you've used is not going to do any counting..... for that you have to do what Dr.Klahn has suggested

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Try using f instead of number when calculating result.