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Difference between revisions of "2011 AMC 10B Problems/Problem 24"

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==Solution==
 
==Solution==
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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>. We see that the nearest fraction bigger than <math>\frac{1}{2}</math> that does not have its denominator over <math>100</math> is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.
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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 the nearest fraction bigger than <math>\frac{1}{2}</math> that does not have its denominator over <math>100</math> 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 14:28, 31 January 2016

Problem

A lattice point in an $xy$-coordinate system is any point $(x, y)$ where both $x$ and $y$ are integers. The graph of $y = mx +2$ passes through no lattice point with $0 < x \le 100$ for all $m$ such that $\frac{1}{2} < m < a$. What is the maximum possible value of $a$?

$\textbf{(A)}\ \frac{51}{101} \qquad\textbf{(B)}\ \frac{50}{99} \qquad\textbf{(C)}\ \frac{51}{100} \qquad\textbf{(D)}\ \frac{52}{101} \qquad\textbf{(E)}\ \frac{13}{25}$

Solution

We see that for the graph of $y=mx+2$ to not pass through any lattice points, the denominator of $m$ must be greater than $100$, or else it would be canceled by some $0<x\le100$ which would make $y$ an integer. We see that the nearest fraction bigger than $\frac{1}{2}$ that does not have its denominator over $100$ is $\boxed{\textbf{(B)}\frac{50}{99}}$.

See Also

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AMC 10 Problems and Solutions

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