# Java solution based on modulus property

• Let's find situation where there is no solution.

1. x is multiple of y
2. y is multiple of x.
3. x is 0 or y is 0
4. x+y>z

This does not mathematically prove that there is no other conditions, but this is enough to pass all the tests, and when I picked x and y that are not multiples of each other I was able to find the solution.
``````public boolean canMeasureWater(int x, int y, int z) {
if (z==x+y || z==x || z==y || z==0)
return true;
if (x==0 || y==0 || x%y ==0 || y%x ==0 || z>x+y)
return false;
return true;
}``````

• x = 4, y = 6, z = 1? should be false right? but returns true in your program i think? correct me if i'm wrong

• Modified to pass 4,6,1 case

``````public boolean canMeasureWater(int x, int y, int z) {
if (z==x+y || z==x || z==y || z==0)
return true;
if (x==0 || y==0 || z>x+y || z%GCF(x,y)!=0)
return false;
return true;
}
private int GCF(int x, int y) {
return y==0?x:GCF(y, x%y);
}``````

• Thanks, using greatest common factor takes care of this case.