Integer Circles 2

Computer Science Level 5

Given the equation of a circle as ${ x }^{ 2 }+{ y }^{ 2 }=1234567$ .How many points inside the the circle have integer coordinates?

This problem is a part of my set The Best of Me

The answer is 3878561.

5 solutions

Drop TheProblem
Oct 25, 2014

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{int conta=0;
for(int x=-1111;x<=1111;x++) 
   for(int y=-1111;y<=1111;y++)
      if(x*x+y*y<1234567)conta++;
cout<<conta;
return 0;
}

$\boxed{conta=3878561}$

Manit Kapoor
Sep 24, 2014

use this easy program(easy for us hard on complier)

include<stdio.h>

void main() { long long int x, y, n1=0, d; for(x=-1200;x<1200;x++){for(y=-1200;y<1200;y++){ d=(x x)+(y y); if(d <= 1234567) { n1++; } } y=0; } printf("\n%lld",n1); }

Swapnil Bhargava
Sep 19, 2014

#include<iostream>
#include<cmath>
#define radius 1111
using namespace std;
int main()
{
    long long int ans=0;
    int i;
    for(i=1;i<=radius;i++)
    {
        ans+=(long long int)floor(sqrt(1234567-(i*i)));
    }
    ans*=4;//for the 4 quadrants
    ans+=(radius*4)+1;//number of integer points on the axes(1111x4) and the origin
    cout<<ans;
}

Kenny Lau
Sep 16, 2014

java code:

public class brilliant201409161915{
    public static void main(String args[]){
        int count = 0;
        for(int x=((int)-Math.sqrt(1234567));x<=((int)Math.sqrt(1234567));x++){
            for(int y=((int)-Math.sqrt(1234567));y<=((int)Math.sqrt(1234567));y++){
                if(x*x+y*y<1234567) count++;
            }
        }
        System.out.print("    "+count+"\n    ");
    }
}

output:

3878561
Press any key to continue . . .

Siddharth G
Aug 28, 2014

#include <iostream>                                                    
#include<cmath>                                                                                    
using namespace std;                                                                                      
int main()                                                                              
 { long long int x,y,z,i,j,k;                     
  cin>>z;
  k=0; 
  for(x=0;x<=z;x++)                                                                       
    {if (x*x<=z && (x+1)*(x+1)>z)  
        y=x;      
     else                                  
         continue;
       }    
       for(i=-y;i<=y;i++)
      {for(j=-y;j<=y;j++)
         {if ((i*i)+(j*j)<=z)
            k=k+1;
        else
            continue;}
      }
 cout<<k; 
 return 0;
}

Non-computer solutions?

Jayakumar Krishnan - 6 years, 9 months ago

And, hey can you explain the logic of this solution ( @Siddharth G ) ...I dont know...C++ and also did you come up with this question on your own?

Jayakumar Krishnan - 6 years, 9 months ago

Yup, this my own ques. However, I started doing this as a number theory question, tried for a week, got frustrated, and did it with a program. I was hoping someone would provide a different approach. Coming to the logic, it is just a (partial) brute force for finding all $x^2+y^2 <= 1234567$

Siddharth G - 6 years, 9 months ago

WoW! NICE2K....How do you know so much C++? How many languages do you know to code in?

Jayakumar Krishnan - 6 years, 9 months ago

@Jayakumar Krishnan – Python too. If you are interested, https://courses.edx.org/courses/MITx/6.00.1x/3T2013/courseware/Week_1/ provides an awesome course!

Siddharth G - 6 years, 9 months ago

@Siddharth G – Uh-oH! it provides an error messages... BTW @Siddharth G -could you provide the name of the course instead,

Jayakumar Krishnan - 6 years, 9 months ago

Integer Circles 2

The answer is 3878561.

5 solutions

include<stdio.h>

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