The maximum product will be something like 3 * 3 * 3 or 3 * 3 * 4 or 3 * 3 * 2 - the power of 3 and the last multiplier depending on a remainder of n devision on 3.

```
public int integerBreak(int n) {
if (n == 2)
return 1;
if (n == 3)
return 2;
int remainder = n % 3;
if (remainder == 0)
return (int) Math.pow(3, (n/3));
if (remainder == 1)
return (int) Math.pow(3, (n/3 - 1))*4;
if (remainder == 2)
return (int) Math.pow(3, (n/3))*2;
return 0;
}
```