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

Latest revision as of 19:45, 19 March 2023

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 $m=\frac{50}{99}$ , $y$ could 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 largest 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 integral value.

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 $\frac{50}{99}.\boxed{\mathrm{(B)}}$

~ Nafer

Solution 5 (MAA.org)

https://www.maa.org/sites/default/files/pdf/CurriculumInspirations/CB095_A-Line-through-Lattice-Points.pdf

@@ Line 5: / Line 5: @@
 <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==
-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>.
+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 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==
+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. By using common denominators, we find that the order of the fractions from smallest to largest is <math>\text{(A), (B), (C), (D), (E)}</math>. We can see that when <math>m=\frac{50}{99}</math>, <math>y</math> could be an integer, so therefore any fraction greater than <math>\frac{50}{99}</math> would not work, as substituting our fraction <math>\frac{50}{99}</math> for <math>m</math> would produce an integer for <math>y</math>. So now we are left with only <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>. But since <math>\frac{51}{101}=\frac{5049}{9999}</math> and <math>\frac{50}{99}=\frac{5050}{9999}</math>, we can be absolutely certain that there isn't a number between <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math> that can reduce to a fraction whose denominator is less than or equal to <math>100</math>. Since we are looking for the maximum value of <math>a</math>, we take the larger of <math>\frac{51}{101}</math> and <math>\frac{50}{99}</math>, which is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.
+==Solution 3==
+We want to find the smallest <math>m</math> such that there will be an integral solution to <math>y=mx+2</math> with <math>0<x\le100</math>. We first test A, but since the denominator has a <math>101</math>, <math>x</math> must be a nonzero multiple of <math>101</math>, but it then will be greater than <math>100</math>. We then test B. <math>y=\frac{50}{99}x+2</math> yields the solution <math>(99,52)</math> which satisfies <math>0<x\le100</math>. Checking the answer choices, we know that the largest possible <math>a</math> must be <math>\frac{50}{99}\implies\boxed{\textbf{(B)}}</math>
+== Solution 4 ==
+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 integral value.
+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>.
+Thus the maximum value of <math>a</math> is <math>\frac{50}{99}.\boxed{\mathrm{(B)}}</math>
+~ Nafer
+== Solution 5 (MAA.org) ==
+https://www.maa.org/sites/default/files/pdf/CurriculumInspirations/CB095_A-Line-through-Lattice-Points.pdf
 ==See Also==
 {{AMC10 box|year=2011|ab=B|num-a=25|num-b=23}}
+[[Category:Introductory Algebra Problems]]
 {{MAA Notice}}

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