Discovering the Perfect Cube Using Factorization Technique

Looking for some guidance on a C/JS program to find perfect cubes using the factorization method below. Any assistance is greatly appreciated.

  • 3 * 3 * 3
  • 3 * 3 * 3
  • 3 * 3 * 3

JavaScript Code

var num=19683;
   var arr=[];
   for(i=2;i<num;i++){
  if(num%i==0){
     arr.push(i);
  }
}

C Code

#include<stdio.h>

int main() {
    int num=19683;
    int a[20];
    int j=0;
    for(int i=2;i<num;i++){
    if(num%i==0){
       a[j]=i;
        j++;

    }

    }
    for(int i=0;i<j;i++){
    printf(" %i", a[i]);
    }
}

Result:

1,3,9,27,81,243,729,2187,6561

javascriptcauthentication

Answer №1

It is not clear what the expected outcome should be, Is this similar to that?

var num=1881365963625;
var arr=[];
var find = "yes";
for(var i = 2; i < num; i++){
  var count = 0;
  while(num % i==0){
     num /= i;
     ++count;
  }
  if(count % 3 != 0){
     find = "no";
     break;
  } else if(count != 0){
     arr.push([i, count]);
  }
}
console.log(find);
if(find == "yes")
  console.log(arr);

Answer №2

Here is an untested fast C-solution:

int isPerfectCube(long int num){
    long int x = pow(num, 1.0/3) + 0.5;
    return (x*x*x == num);
}

While this solution is quick to implement and relies on the pow() function for heavy lifting, it does have a downside due to its use of floating point arithmetic.

To ensure accuracy of pow(num, 1.0/3), it must be closer to the correct integer than any other integer when num is a perfect cube. If not, the algorithm will fail. The addition of + 0.5 rounds this value to the nearest integer upon truncation from double to int.

A potentially faster and safer alternative may involve developing your own iterative algorithm tailored to this specific task.

Answer №3

To determine if the given number is a perfect cube or not

 var num = 19683, c=0;
    for(var i=1; i<=num; i++){
      if((num%i===0)&&(i*i*i === num)){
        c = c+1;
      }
    }
    if(c > 0) {
     console.log('Yes')
    } else console.log('No');

This method might be more efficient.

var num = 100000000, c=0, arr=[];
for(var i=1; i<(num/2)+1; i++){
  if(num%i===0){

    arr.push(i);
  }
}
for(var i=0; i< arr.length; i++) {
   var x = arr[i];
   if(x*x*x === num) c = c+1;
}
if(c > 0) {
 console.log('Yes')
} else console.log('No')

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