﻿ Java - Sum any two cubes between 1 and n in two or more ways

# Java: Find any number between 1 and n that can be expressed as the sum of two cubes in two (or more) different ways

## Java Numbers: Exercise-21 with Solution

Write a Java program to find any number between 1 and n that can be expressed as the sum of two cubes in two (or more) different ways.

//http://introcs.cs.princeton.edu/java/13flow/Ramanujan.java.html
Here are some examples of Ramanujan numbers :
1729 = 1^3 + 12^3 = 9^3 + 10^3
* 10000
1729 = 1^3 + 12^3 = 9^3 + 10^3
4104 = 2^3 + 16^3 = 9^3 + 15^3
* 100000
1729 = 1^3 + 12^3 = 9^3 + 10^3
4104 = 2^3 + 16^3 = 9^3 + 15^3
13832 = 2^3 + 24^3 = 18^3 + 20^3
39312 = 2^3 + 34^3 = 15^3 + 33^3
46683 = 3^3 + 36^3 = 27^3 + 30^3
32832 = 4^3 + 32^3 = 18^3 + 30^3
40033 = 9^3 + 34^3 = 16^3 + 33^3
20683 = 10^3 + 27^3 = 19^3 + 24^3
65728 = 12^3 + 40^3 = 31^3 + 33^3
64232 = 17^3 + 39^3 = 26^3 + 36^3

Sample Solution:

Java Code:

``````import java.util.Scanner;
public class Example21  {

public static void main(String[] args) {

int n = 100000;
// for each a, b, c, d, check whether a^3 + b^3 = c^3 + d^3
for (int a = 1; a <= n; a++) {
int a3 = a*a*a;
if (a3 > n) break;

for (int b = a; b <= n; b++) {
int b3 = b*b*b;
if (a3 + b3 > n) break;

for (int c = a + 1; c <= n; c++) {
int c3 = c*c*c;
if (c3 > a3 + b3) break;

for (int d = c; d <= n; d++) {
int d3 = d*d*d;
if (c3 + d3 > a3 + b3) break;

if (c3 + d3 == a3 + b3) {
System.out.print((a3+b3) + " = ");
System.out.print(a + "^3 + " + b + "^3 = ");
System.out.print(c + "^3 + " + d + "^3");
System.out.println();
}
}
}
}
}
}
}
```
```

Sample Output:

```1729 = 1^3 + 12^3 = 9^3 + 10^3
4104 = 2^3 + 16^3 = 9^3 + 15^3
13832 = 2^3 + 24^3 = 18^3 + 20^3
39312 = 2^3 + 34^3 = 15^3 + 33^3
46683 = 3^3 + 36^3 = 27^3 + 30^3
32832 = 4^3 + 32^3 = 18^3 + 30^3
40033 = 9^3 + 34^3 = 16^3 + 33^3
20683 = 10^3 + 27^3 = 19^3 + 24^3
65728 = 12^3 + 40^3 = 31^3 + 33^3
64232 = 17^3 + 39^3 = 26^3 + 36^3
```

Flowchart: Java Code Editor:

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## Java: Tips of the Day

IsPowerOfTwo

Checks if a value is positive power of two.

To understand how it works let's assume we made a call IsPowerOfTwo(4).

As value is greater than 0, so right side of the && operator will be evaluated.

The result of (~value + 1) is equal to value itself. ~100 + 001 => 011 + 001 => 100. This is equal to value.

The result of (value & value) is value. 100 & 100 => 100.

This will value the expression to true as value is equal to value.

```public static boolean isPowerOfTwo(final int value) {
return value > 0 && ((value & (~value + 1)) == value);
}
```

Ref: https://bit.ly/3sA5d4I

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