Asked by nics
if integers x and y satisfy 2008x=16y , the smallest possible value of x+y is______.
Answers
Answered by
Steve
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.
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.
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