Difference between revisions of "2011 AMC 10B Problems/Problem 24"
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== Solution 1== | == Solution 1== | ||
| − | For <math>y=mx+2</math> to not pass through any lattice points with <math>0<x\leq 100</math> is the same as saying that <math>mx\notin\mathbb Z</math> for <math>x\in\{1,2,\dots,100\}</math>, or in other words, <math>m</math> is not expressible as a ratio of positive integers <math>s/t</math> with <math>t\leq 100</math>. Hence the maximum possible value of <math>a</math> is the first real number after 1/2 that is so expressible. | + | For <math>y=mx+2</math> to not pass through any lattice points with <math>0<x\leq 100</math> is the same as saying that <math>mx\notin\mathbb Z</math> for <math>x\in\{1,2,\dots,100\}</math>, or in other words, <math>m</math> is not expressible as a ratio of positive integers <math>s/t</math> with <math>t\leq 100</math>. Hence the maximum possible value of <math>a</math> is the first real number after <math>1/2</math> that is so expressible. |
| − | For each <math>d=2,\dots,100</math>, the smallest | + | For each <math>d=2,\dots,100</math>, the smallest multiple of <math>1/d</math> which exceeds <math>1/2</math> is <math>1,\frac23,\frac34,\frac35,\dots,\frac{50}{98},\frac{50}{99},\frac{51}{100}</math> respectively, and the smallest of these is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>. |
==Solution 2== | ==Solution 2== | ||
Revision as of 04:23, 22 December 2018
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 . We know that MAA orders the answers in ascending order, so therefore 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
