Difference between revisions of "2011 AMC 10B Problems/Problem 24"
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==Solution 3== | ==Solution 3== | ||
− | We want to find the smallest <math>m</math> such that there will be an integral solution to <math>y=mx+2</math> with <math>0<x\le100</math>. We first test A, but since the denominator has a <math>101</math>, <math>x</math> must be a nonzero multiple of <math>101</math>, but it then will be greater than <math>100</math>. We then test B. <math>y=\frac{50}{99}x+2</math> yields the solution <math>(99,52)</math> which satisfies <math>0<x\le100</math>. | + | We want to find the smallest <math>m</math> such that there will be an integral solution to <math>y=mx+2</math> with <math>0<x\le100</math>. We first test A, but since the denominator has a <math>101</math>, <math>x</math> must be a nonzero multiple of <math>101</math>, but it then will be greater than <math>100</math>. We then test B. <math>y=\frac{50}{99}x+2</math> yields the solution <math>(99,52)</math> which satisfies <math>0<x\le100</math>. Checking the answer choices, we know that the smallest possible <math>a</math> must be <math>\frac{50}{99}\implies\boxed{\textbf{(B)}}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}} | {{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:15, 26 December 2018
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
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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