# 2011 AMC 10B Problems/Problem 24

## 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 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 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. 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. 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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 