2007 AIME II Problems/Problem 5

Problem

The graph of the equation $9x+223y=2007$ is drawn on graph paper with each square representing one unit in each direction. How many of the $1$ by $1$ graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant?

Solution

Solution 1

Count the number of each squares in each row of the triangle. The intercepts of the line are $(223,0),\ (9,0)$ .

In the top row, there clearly are no squares that can be formed. In the second row, we see that the line $y = 8$ gives a $x$ value of $\frac{2007 - 8(223)}{9} = 24 \frac 79$ , which means that $\lfloor 24 \frac 79\rfloor = 24$ unit squares can fit in that row. In general, there are

$\displaystyle\sum_{i=0}^{8} \lfloor \frac{223i}{9} \rfloor$

triangles. Since $\lfloor \frac{223}{9} \rfloor = 24$ , we see that there are more than $24(0 + 1 + \ldots + 8) = 24(\frac{8 \times 9}{2}) = 864$ triangles. Now, count the fractional parts. $\lfloor \frac{0}{9} \rfloor = 0, \lfloor \frac{7}{9} \rfloor = 0, \lfloor \frac{14}{9} \rfloor = 1,$ $\lfloor \frac{21}{9} \rfloor = 2, \lfloor \frac{28}{9} \rfloor = 3, \lfloor \frac{35}{9} \rfloor = 3,$ $\lfloor \frac{42}{9} \rfloor = 4, \lfloor \frac{49}{9} \rfloor = 5, \lfloor \frac{56}{9} \rfloor = 6$ . Adding them up, we get $864 + 1 + 2 + 3 + 3 + 4 + 5 + 6 = 888 \displaystyle$ .

Solution 2

From Pick's Theorem, $\frac{2007}{2}=\frac{233}{2}-\frac{2}{2}+\frac{2I}{2}$ . In other words, $2I=1776$ and I is $\displaystyle 888$ .

2007 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2007 AIME II Problems/Problem 5

Problem

Contents

Solution

Solution 1

Solution 2

See also