Difference between revisions of "2011 AMC 10B Problems/Problem 24"
(→Solution 4) |
(→Solution 4) |
||
Line 18: | Line 18: | ||
== Solution 4 == | == Solution 4 == | ||
− | Notice that for <math>\frac{1}{2}x=\frac{50}{100}x</math>, <math>x=99</math> is one of the integral values of <math>x</math> such that the value of <math>\frac{50}{100}x</math> is the closest to its next lattice point, or the next integral value of <math>y</math>. | + | Notice that for <math>y=\frac{1}{2}x+2=\frac{50}{100}x+2</math>, <math>x=99</math> is one of the integral values of <math>x</math> such that the value of <math>\frac{50}{100}x</math> is the closest to its next lattice point, or the next integral value of <math>y</math>. |
+ | |||
+ | Thus the maximum value for <math>a</math> is the value of <math>m</math> when the equation <math>y=99m+2</math> goes through its next lattice point, which occurs when <math>m=\frac{b}{99}</math> for some positive integer <math>b</math>. | ||
+ | |||
+ | Finding the common denominator, we have <cmath>\frac{50}{100}=\frac{4950}{9900}, \frac{b}{99}=\frac{100b}{9900}</cmath> The smallest value for <math>b</math> such that <math>100b>4950</math> is <math>b=50</math>. | ||
+ | |||
+ | Thus the maximum value of <math>a</math> is $\frac{50}{99}. |
Revision as of 11:19, 6 August 2019
Contents
[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}.