2011 AMC 10B Problems/Problem 24: Difference between revisions

Latest revision as of 12:29, 4 November 2025

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 1.1

Note that finding the next possible slope $m$ greater than $\frac{1}{2}$ is the same as finding the next Farey fraction of order $100$ after $\frac{1}{2}$ . Hence, we can use the Farey mediant theorem (https://en.wikipedia.org/wiki/Farey_sequence#Farey_neighbours), which states that if $\frac{a}{b}<\frac{c}{d}$ are neighboring Farey fractions of order $n$ , then $bc-ad=1$ where $b, d \leq n$ .

In this case, we have $a=1$ , $b=2$ , and so $2c-d=1$ . We try $d=100$ , but that doesn't give integer $c$ . We try $d=99$ , which gives us $c=50$ so the upper bound is $\boxed{\textbf{(B)}\frac{50}{99}}$ .

~Math-lover1

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 9: / Line 9: @@
 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 1.1===
+Note that finding the next possible slope <math>m</math> greater than <math>\frac{1}{2}</math> is the same as finding the next Farey fraction of order <math>100</math> after <math>\frac{1}{2}</math>. Hence, we can use the Farey mediant theorem (https://en.wikipedia.org/wiki/Farey_sequence#Farey_neighbours), which states that if <math>\frac{a}{b}<\frac{c}{d}</math> are neighboring Farey fractions of order <math>n</math>, then <math>bc-ad=1</math> where <math>b, d \leq n</math>.
+In this case, we have <math>a=1</math>, <math>b=2</math>, and so <math>2c-d=1</math>. We try <math>d=100</math>, but that doesn't give integer <math>c</math>. We try <math>d=99</math>, which gives us <math>c=50</math> so the upper bound is <math>\boxed{\textbf{(B)}\frac{50}{99}}</math>.
+~Math-lover1
 ==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> would 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>.
+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 smallest possible <math>a</math> must be <math>\frac{50}{99}\implies\boxed{\textbf{(B)}}</math>
+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 ==
@@ Line 30: / Line 36: @@
 ~ Nafer
+== Solution 5 (MAA.org) ==
+https://www.maa.org/sites/default/files/pdf/CurriculumInspirations/CB095_A-Line-through-Lattice-Points.pdf
 ==See Also==

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