2011 AMC 10B Problems/Problem 24
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[hide]Problem
A lattice point in an -coordinate system is any point where both and are integers. The graph of passes through no lattice point with for all such that . What is the maximum possible value of ?
Solution 1
For to not pass through any lattice points with is the same as saying that for , or in other words, is not expressible as a ratio of positive integers with . Hence the maximum possible value of is the first real number after that is so expressible.
For each , the smallest multiple of which exceeds is respectively, and the smallest of these is .
Solution 2
We see that for the graph of to not pass through any lattice points, the denominator of must be greater than , or else it would be canceled by some which would make an integer. By using common denominators, we find that the order of the fractions from smallest to largest is . We can see that when , would be an integer, so therefore any fraction greater than would not work, as substituting our fraction for would produce an integer for . So now we are left with only and . But since and , we can be absolutely certain that there isn't a number between and that can reduce to a fraction whose denominator is less than or equal to . Since we are looking for the maximum value of , we take the larger of and , which is .
Solution 3
We want to find the smallest such that there will be an integral solution to with . We first test A, but since the denominator has a , must be a nonzero multiple of , but it then will be greater than . We then test B. yields the solution which satisfies . Checking the answer choices, we know that the smallest possible must be
Solution 4
Notice that for , is one of the integral values of such that the value of is the closest to its next lattice point, or the next integral value of .
Thus the maximum value for is the value of when the equation goes through its next lattice point, which occurs when for some positive integer .
Finding the common denominator, we have The smallest value for such that is .
Thus the maximum value of is $\frac{50}{99}.