if integers x and y satisfy 2008x=16y , the smallest possible value of x+y is______.

Answer 1

Using positive integers,
since 2008=8*251 and 16 = 8*2,
251x = 2y

x=2, y=251 gives x+y=253

If you allow zero, then 0+0=0 is the minimum.

now, if you allow negative integers, then there is no minimum value.