2011 AMC 12B Problem 19

Question

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 \lt x \le 100

for all

m

such that

\dfrac{1}{2} \lt m \lt a.

What is the maximum possible value of

a?

\dfrac{51}{101}

\dfrac{50}{99}

\dfrac{51}{100}

\dfrac{52}{101}

\dfrac{13}{25}

Answer: B

Concepts:lattice point slope

Difficulty rating: 2090

Small hint:

For each

x,

find the smallest slope

\gt\dfrac12

from

(0,2)

that reaches a lattice point at that

x

Big hint:

Compare the minimum slopes for even

x

and odd

x;

the tighter bound wins

Solution:

For

0\lt x\le100,

the nearest lattice point above the line

y=\tfrac12x+2

is

\left(x,\tfrac12x+3\right)

if

x

is even and

\left(x,\tfrac12x+\tfrac52\right)

if

x

is odd.

The slope from

(0,2)

to that point is

\dfrac12+\dfrac1x

for even

x

and

\dfrac12+\dfrac{1}{2x}

for odd

x.

The minimum such slope is

\dfrac{51}{100}

for even

x

and

\dfrac{50}{99}

for odd

x.

Since

\dfrac{50}{99}\lt\dfrac{51}{100},

the line avoids all these lattice points exactly when

\dfrac12\lt m\lt\dfrac{50}{99},

so the maximum is

a=\dfrac{50}{99}.

Thus, the correct answer is B.