Difference between revisions of "2011 AMC 12B Problems/Problem 19"
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It is very easy to see that the <math>+2 </math> in the graph does not impact whether it passes through lattice. | It is very easy to see that the <math>+2 </math> in the graph does not impact whether it passes through lattice. | ||
− | We need to make sure that <math>m</math> cannot be in the form of <math>\frac{a}{b}</math> for <math>1\le b\le 100</math>, otherwise the graph <math>y= mx</math> passes through lattice point at <math>x = b</math>. We only need to worry about <math>\frac{a}{b}</math> very close to <math>\frac{1}{2}</math>, <math>\frac{m+1}{2m+1}</math>, <math>\frac{m+1}{2m}</math> will be the only case we need to worry about and we want the minimum of those, clearly for <math>1\le b\le 100</math>, the smallest is <math>\frac{50}{99} </math>, so answer is <math>\boxed{\frac{50}{99}\ \(\textbf{(B)}}</math> | + | We need to make sure that <math>m</math> cannot be in the form of <math>\frac{a}{b}</math> for <math>1\le b\le 100</math>, otherwise the graph <math>y= mx</math> passes through lattice point at <math>x = b</math>. We only need to worry about <math>\frac{a}{b}</math> very close to <math>\frac{1}{2}</math>, <math>\frac{m+1}{2m+1}</math>, <math>\frac{m+1}{2m}</math> will be the only case we need to worry about and we want the minimum of those, clearly for <math>1\le b\le 100</math>, the smallest is <math>\frac{50}{99} </math>, so answer is <math>\boxed{\frac{50}{99} \textbf{(B)}}</math> |
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+ | ==Solution 2== | ||
+ | Like in the first solution, we note that the <math>+2</math> does not affect our answer. Thus, we can ignore it and consider an equivalent problem with the line <math>y = mx</math>. | ||
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+ | A line with a slope of <math>\frac{1}{2}</math> passes through (among other lattice points) the lattice point <math>(100,50)</math>. As the slope of the line increases from <math>\frac{1}{2}</math>, the first lattice point it hits is at <math>(99, 50)</math>, the slope of that line being <math>\frac{50}{99}</math> . So the answer is <math>\boxed{\textbf{(B)}}</math> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=18|num-a=20|ab=B}} | {{AMC12 box|year=2011|num-b=18|num-a=20|ab=B}} | ||
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+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:49, 22 August 2020
Contents
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
It is very easy to see that the in the graph does not impact whether it passes through lattice.
We need to make sure that cannot be in the form of for , otherwise the graph passes through lattice point at . We only need to worry about very close to , , will be the only case we need to worry about and we want the minimum of those, clearly for , the smallest is , so answer is
Solution 2
Like in the first solution, we note that the does not affect our answer. Thus, we can ignore it and consider an equivalent problem with the line .
A line with a slope of passes through (among other lattice points) the lattice point . As the slope of the line increases from , the first lattice point it hits is at , the slope of that line being . So the answer is
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 12 Problems and Solutions |
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