largest factor of a number

[salmanasghar12]

here is the code for Displaying Factors of a Number

  #include <iostream>
using namespace std;
int main()
{
  int n,i,a;
  cout << "Enter a positive integer: ";
  cin >> n;
  cout << "Factors of " << n << " are: " << endl;  
  for(i=1;i<=n;++i)
  {
      if(n%i==0)
         cout << i << endl;
  }
  for(i=1;i<=n;++i)
  
  
  return 0;
}

can anyone guess how to just print out largest factor...????

[shadder]

Hi,

can anyone guess how to just print out largest factor...????

technically the largest factor is the number itself so you are better off printing the number itself

so cout<<n will do the magic!

but if you are trying to find out the largest factor (apart from) the number itself you may use a array and then print the second last digit, and if there are only 2 digit then the number is prime and then it won't have largest factor (well except the number itself)

also, it would be great if you could show us your code/attempt...

[ajputnam]

Working off of Programmer007 since factors are always in pairs you can stop at the second factor.

example #12
factors: 1 2 3 4 6 12
stop at 2 since you can do : 2*x=12 => 12/2 =6

[shadder]

That's a great approach!

[Yawzheek]

Well since a factor (other than the number itself) can never really be more than half the value of an integer, say n, you could test until another integer i is <= i/n and i%n == 0, I would think. Surely there are better ways, but that just sprang to mind quickly. Store them in a vector v and print v[v.size() - 1]

EDIT: Though the vector is probably overkill. You could probably just use a local variable and rewrite it as needed, then print that.

int greatest_factor{0};
for(size_t i = 0; i <= n/2; ++i) if (i % n == 0) greatest_factor = i;
cout << "The greatest factor of the number " << n << " is " << greatest_factor << endl;

Or something like that.

[salmanasghar12]

i was talking about largest factor of a number apart from the number itself and question arised when we see that if a number is even logic number/2 perfectly works but in case of odd number it can't work so required to make a good logic

#include <iostream>
using namespace std;
int main()
{
	int i,n;
	
  cout << "Enter a positive integer: ";
  cin >> n;
  cout << "Factors of " << n << " are: " << endl;  
  for(i=1;i<=n;++i)
  {
      if(n%i==0)
         cout << i << endl;
  }
 cout <<"\n largest factor of number is " <<endl;
 cout <<n/2;
  
/* problem is that when you have to
 find greatest factor of an odd number
this logic n/2 does not work , there
 should be something else */
  
  return 0;
}

output is like

if i enter 8
it gives factor are 1,2,4,8
and
largest factor of number is 4
but in case of odd no. like 9
it gives factors are 1,3,9
and 
largest factor is 4 //this is error..

[Yawzheek]

I'm... not sure what you're asking for? I just ran it in Shell, and I received:

Enter a positive integer: 9
Factors of 9 are:
1
3
9

And nothing more, which is still correct. Division by 2 - even taking into account modulo arithmetic - isn't going to change the result for odd numbers, because 4.5? Not really an acceptable answer. Even running it with an even number:

Enter a positive integer: 8
Factors of 8 are:
1
2
4
8

You'll always receive the original number. If you ONLY want the LARGEST factor that ISN'T the original number (because otherwise n == n, yes?) Well...

#include <iostream>

using namespace std;

int main()
{
	cout << "Enter a positive number: ";
	int n{ 0 };
	cin >> n;
	int largest_factor{ 0 };
	for (unsigned i = 1; i <= n / 2; ++i) if (n % i == 0) largest_factor = i;
	cout << "The largest factor of " << n << " == " << largest_factor << endl;
        return 0;
}

Is that what you're looking for?

EDIT: Removed an #include of vector - copy/pasted from VS from something I was working on previously. Sorry.

[salmanasghar12]

@yawzheek your code didn't worked on Codeblocks and visual studio 2015 update 2
it says something like

[Warning] extended initializer lists only available with -std=c++11 or -std=gnu++11

and for all who see i want to make a program that can make largest factor of an odd number like 9

[Duthomhas]

At this point I believe you are just asking someone to do your thinking for you; user ajputnam has already given you the simplest, most correct answer: http://www.cplusplus.com/forum/beginner/193443/#msg930540

It takes a very simple loop with a break statement just like finding all the factors would; the only difference is stopping at the right spot.