Difference between revisions of "2011 AMC 12B Problems/Problem 19"

(Created page with '==Problem== A lattice point in an <math>xy</math>-coordinate system is any point <math>(x, y)</math> where both <math>x</math> and <math>y</math> are integers. The graph of <math…')
 
(Solution)
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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 (B)
+
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>
 
 
  
 
== 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}}

Revision as of 03:40, 9 March 2012

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 $0 < x \leq 100$ 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

Answer: (B)

It is very easy to see that the $+2$ in the graph does not impact whether it passes through lattice.

We need to make sure that $m$ cannot be in the form of $\frac{a}{b}$ for $1\le b\le 100$, otherwise the graph $y= mx$ passes through lattice point at $x = b$. We only need to worry about $\frac{a}{b}$ very close to $\frac{1}{2}$, $\frac{m+1}{2m+1}$, $\frac{m+1}{2m}$ will be the only case we need to worry about and we want the minimum of those, clearly for $1\le b\le 100$, the smallest is $\frac{50}{99}$, so answer is $\boxed{\frac{50}{99}\ \(\textbf{(B)}}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
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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