Question

C program

//In this assignment, we will find the smallest positive integer that
// can be expressed as a sum of two positive cube numbers in two distinct ways.
// More specifically, we want to find the smallest n such that n == i1*i1*i1 + j1*j1*j1,
// n == i2*i2*i2 + j2*j2*j2, and (i1, j1) and (i2, j2) are not the same in the sense that
// not only (i1, j1) not euqal to (i2, j2), but also (i1, j1) not equal to (j2, i2).
// We practice using loops in this assignment.
// Also, we practice how to do an early exit in a loop. We exit the loop once we find the smallest n.
// Once we are done, we will be delighted surprised to find out that this number already played a big role in
// our computer science study.
#include <stdio.h>
int main()
{
   int i, j, n;

   //We assume the smallest such n is no more than 1000000
   for(n=1; n<=1000000; n++)
   {
       //Use count to record the number of different ways of summing two positive cube numbers to n
       int count = 0;
       //TO DO
       //Fill in code below

   }

   //Do not change code below
printf("%d is the solution:\n", n);
   for(i=1; i<=100; i++)
       for(j = i; j<= 100; j++)
           {
               if (i*i*i + j*j*j == n)
               {
                   printf("%d * %d * %d + %d * %d * %d = %d\n", i, i, i, j, j, j, n);
}
           }

   return 0;
   //Do not try to hard code the solution, that is not the way to learn.
}

Answer 1

Here is the code:

#include <stdio.h>
#include <math.h>

int main()
{
int i, j, n;

//We assume the smallest such n is no more than 1000000
for( n=1; n<=1000000; n++)
{
//Use count to record the number of different ways of summing two positive cube numbers to n
int count = 0;

  // find cube root of n using cbrt

   for (int i = 1; i <= cbrt(n); i++)
for (int j = i + 1; j <= cbrt(n); j++)
if (i*i*i + j*j*j == n)
count++;

if (count == 2)
  
break;
  


}
     
  

//Do not change code below
printf("%d is the solution:\n", n);
for(i=1; i<=100; i++)
{
for(j = i; j<= 100; j++)
{
if (i*i*i + j*j*j == n)
{
printf("%d * %d * %d + %d * %d * %d = %d\n", i, i, i, j, j, j, n);
}
} }

return 0;
//Do not try to hard code the solution, that is not the way to learn.
}

OUTPUT:

1729 is the solution:
1 * 1 * 1 + 12 * 12 * 12 = 1729
9 * 9 * 9 + 10 * 10 * 10 = 1729

Logic is:

If a number X has pairs like (a,b) and (c,d) then as X<N, all the numbers a,b,c,d must be less than cube root of N