Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 4)
(→‎Solution 4)
Line 19: Line 19:
  
 
Notice that for <math>y=\frac{1}{2}x+2=\frac{50}{100}x+2</math>, <math>x=99</math> is one of the integral values of <math>x</math> such that the value of <math>\frac{50}{100}x</math> is the closest to its next lattice point, or the next integral value of <math>y</math>.
 
Notice that for <math>y=\frac{1}{2}x+2=\frac{50}{100}x+2</math>, <math>x=99</math> is one of the integral values of <math>x</math> such that the value of <math>\frac{50}{100}x</math> is the closest to its next lattice point, or the next integral value of <math>y</math>.
 +
  
 
Thus the maximum value for <math>a</math> is the value of <math>m</math> when the equation <math>y=99m+2</math> goes through its next lattice point, which occurs when <math>m=\frac{b}{99}</math> for some positive integer <math>b</math>.
 
Thus the maximum value for <math>a</math> is the value of <math>m</math> when the equation <math>y=99m+2</math> goes through its next lattice point, which occurs when <math>m=\frac{b}{99}</math> for some positive integer <math>b</math>.
 +
  
 
Finding the common denominator, we have <cmath>\frac{50}{100}=\frac{4950}{9900}, \frac{b}{99}=\frac{100b}{9900}</cmath> Since <math>a>\frac{1}{2}</math>, the smallest value for <math>b</math> such that <math>100b>4950</math> is <math>b=50</math>.  
 
Finding the common denominator, we have <cmath>\frac{50}{100}=\frac{4950}{9900}, \frac{b}{99}=\frac{100b}{9900}</cmath> Since <math>a>\frac{1}{2}</math>, the smallest value for <math>b</math> such that <math>100b>4950</math> is <math>b=50</math>.  
  
−
Thus the maximum value of <math>a</math> is $\frac{50}{99}.
+
 
 +
Thus the maximum value of <math>a</math> is <math>\frac{50}{99}.\boxed{mathrm{(B)}}</math>

Revision as of 11:21, 6 August 2019

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 1

For $y=mx+2$ to not pass through any lattice points with $0<x\leq 100$ is the same as saying that $mx\notin\mathbb Z$ for $x\in\{1,2,\dots,100\}$, or in other words, $m$ is not expressible as a ratio of positive integers $s/t$ with $t\leq 100$. Hence the maximum possible value of $a$ is the first real number after $1/2$ that is so expressible.

For each $d=2,\dots,100$, the smallest multiple of $1/d$ which exceeds $1/2$ is $1,\frac23,\frac34,\frac35,\dots,\frac{50}{98},\frac{50}{99},\frac{51}{100}$ respectively, and the smallest of these is $\boxed{\textbf{(B)}\frac{50}{99}}$.

Solution 2

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. By using common denominators, we find that the order of the fractions from smallest to largest is $\text{(A), (B), (C), (D), (E)}$. We can see that when $m=\frac{50}{99}$, $y$ would be an integer, so therefore any fraction greater than $\frac{50}{99}$ would not work, as substituting our fraction $\frac{50}{99}$ for $m$ would produce an integer for $y$. So now we are left with only $\frac{51}{101}$ and $\frac{50}{99}$. But since $\frac{51}{101}=\frac{5049}{9999}$ and $\frac{50}{99}=\frac{5050}{9999}$, we can be absolutely certain that there isn't a number between $\frac{51}{101}$ and $\frac{50}{99}$ that can reduce to a fraction whose denominator is less than or equal to $100$. Since we are looking for the maximum value of $a$, we take the larger of $\frac{51}{101}$ and $\frac{50}{99}$, which is $\boxed{\textbf{(B)}\frac{50}{99}}$.

Solution 3

We want to find the smallest $m$ such that there will be an integral solution to $y=mx+2$ with $0<x\le100$. We first test A, but since the denominator has a $101$, $x$ must be a nonzero multiple of $101$, but it then will be greater than $100$. We then test B. $y=\frac{50}{99}x+2$ yields the solution $(99,52)$ which satisfies $0<x\le100$. Checking the answer choices, we know that the smallest possible $a$ must be $\frac{50}{99}\implies\boxed{\textbf{(B)}}$

Solution 4

Notice that for $y=\frac{1}{2}x+2=\frac{50}{100}x+2$, $x=99$ is one of the integral values of $x$ such that the value of $\frac{50}{100}x$ is the closest to its next lattice point, or the next integral value of $y$.


Thus the maximum value for $a$ is the value of $m$ when the equation $y=99m+2$ goes through its next lattice point, which occurs when $m=\frac{b}{99}$ for some positive integer $b$.


Finding the common denominator, we have \[\frac{50}{100}=\frac{4950}{9900}, \frac{b}{99}=\frac{100b}{9900}\] Since $a>\frac{1}{2}$, the smallest value for $b$ such that $100b>4950$ is $b=50$.


Thus the maximum value of $a$ is $\frac{50}{99}.\boxed{mathrm{(B)}}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_24&oldid=108313"