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

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<math>\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}</math>
 
<math>\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}</math>
  
−
==Solution==
+
==Solution 1==
−
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 the 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 the 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{n+1}{2n+1}</math>, <math>\frac{n+1}{2n}</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> (In other words we are trying to find the smallest <math>m>\frac{1}{2}</math> such that <math>b\le 100</math>.)
 +
 
 +
==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>.
 +
 
 +
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>
 +
 
 +
 
 +
==Solution 3==
 +
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>.
 +
 
 +
We will start with the first answer choice. Notice how it is impossible for a line with slope <math>\frac{51}{101}</math> to pass through a lattice point when <math>0 < x \leq 100</math>.
 +
 
 +
However, if the slope is <math>\frac{50}{99}</math>, then the line will pass through the point <math>(99, 50)</math>. This sets the upper bound for <math>a</math> as <math>\boxed{\textbf{(B)}}</math>.
 +
 
 +
~xHypotenuse
  
 
== 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}}
 +
 +
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:02, 13 July 2024

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 1

It is very easy to see that the $+2$ in the graph does not impact whether it passes through the 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 the lattice point at $x = b$. We only need to worry about $\frac{a}{b}$ very close to $\frac{1}{2}$, $\frac{n+1}{2n+1}$, $\frac{n+1}{2n}$ 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 $\boxed{\frac{50}{99} \textbf{(B)}}$ (In other words we are trying to find the smallest $m>\frac{1}{2}$ such that $b\le 100$.)

Solution 2

Like in the first solution, we note that the $+2$ does not affect our answer. Thus, we can ignore it and consider an equivalent problem with the line $y = mx$.

A line with a slope of $\frac{1}{2}$ passes through (among other lattice points) the lattice point $(100,50)$. As the slope of the line increases from $\frac{1}{2}$, the first lattice point it hits is at $(99, 50)$, the slope of that line being $\frac{50}{99}$ . So the answer is $\boxed{\textbf{(B)}}$


Solution 3

Like in the first solution, we note that the $+2$ does not affect our answer. Thus, we can ignore it and consider an equivalent problem with the line $y = mx$.

We will start with the first answer choice. Notice how it is impossible for a line with slope $\frac{51}{101}$ to pass through a lattice point when $0 < x \leq 100$.

However, if the slope is $\frac{50}{99}$, then the line will pass through the point $(99, 50)$. This sets the upper bound for $a$ as $\boxed{\textbf{(B)}}$.

~xHypotenuse

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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