hihi, all - i think the answer is no for any ratio - irrational ones are easy, but it seems that you could easily get fractions smaller and larger than a given a/b c/d < a/b < (c+1)/(d+1) so that the value would never be exact Thane Plambeck asked Say you're shooting baskets and you keep track of your record as you shoot them, for example 0 for 1 1 for 2 1 for 3 1 for 4 1 for 5 2 for 6 etc, you get the idea Someone specifies a "target ratio" (I think it was 80% in the problem statement) and you're told that during the record of shots, the shooter was below the target ratio, but later was better than the target ratio. Is there necessarily a moment at which the shooter must have shot exactly that ratio up to that point? cal again - ok, so i' ve changed my mind - the above inequalities require bc < ad a(d+1) < b(c+1) which are equivalent to 0 < ad - bc < b - a so if b=a-1, there is no solution for c and d - if a and b have a common divisor, it is also a divisor of ad-bc, so we may assume that a and b are relatively prime so i now think that the answer is yes for 80% = 4/5 and any other a/(a+1) fraction, but no for any others more later, cal