Difference between revisions of "2011 AMC 10B Problems/Problem 24"
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==Solution== | ==Solution== | ||
| − | We see that for the graph of <math>y=mx+2</math> to not pass through any lattice points, the denominator of <math>m</math> must be greater than <math>100</math>, or else it would be canceled by some <math>0<x\le100</math> which would make <math>y</math> an integer. We see that | + | We see that for the graph of <math>y=mx+2</math> to not pass through any lattice points, the denominator of <math>m</math> must be greater than <math>100</math>, or else it would be canceled by some <math>0<x\le100</math> which would make <math>y</math> an integer. By using common denominators, we find that the order of the fractions from smallest to largest is <math>(A), (B), (C), (D), (E)</math>. We can see that when <math>x=\frac{50}{99}</math>, <math>y</math> would be an integer, so therefore any fraction greater than <math>\frac{50}{99}</math> would not work, as substituting <math>\frac{50}{99}</math> for <math>m</math> would produce an integer for <math>y</math>. So now we are left with only <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>. But because <math>\frac{51}{101}=\frac{5049}{9999}</math> and <math>\frac{50}{99}=\frac{5050}{9999}</math>, we can be certain that there is no number between <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math> that can reduce to a fraction whose denominator is less than or equal to <math>100</math>. Since we are looking for the maximum value of <math>a</math>, we take the larger of <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>, which is <math>\boxed{\textbf{(B)}\frac{50}{99}}</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 14:51, 31 January 2016
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
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 for would produce an integer for . So now we are left with only 
and . But because 
and 
, we can be certain that there is no 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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
